Practical – Lab reports 1-8


Introduction to Microbiology (University College London)

Laboratory report: experiment 1 Culture medium preparation

Text Box: Introduction
Briefly describe what a culture medium is, why it needs to be sterile and what the aims of the experiment were.

A culture medium is a liquid or gelatinous substance that contains essential nutrients, to cultivate target microorganisms or tissues, for further purposes. A culture medium must be sterilised before use so that no unwanted microorganisms grow, which may contaminate the growing sample.

The aim of this experiment is to demonstrate that culture media are contaminated naturally (i.e. untreated). Two possible sterilisation techniques – autoclaving & filtration are also tested.

Text Box: Discussion
Discuss what your results tell you about the autoclaving and filtration methods. Discuss the relative merits of a) doing nothing; b) autoclaving; c) filtering as strategies for preparing sterile media. Discuss possible reasons for the variation seen in the class results. Can you suggest alternative techniques for media sterilization?

The absence of growth of bacteria in the autoclaved and filtered flasks indicated that these  two methods maybe effective methods of sterilisation (removing/killing all microbes). Growth in the untreated flask also indicated that natural culture media is contaminated.

The unsuccessful cases in autoclaving & filtering may due to several reasons. Firstly, nutrient broth maybe contaminated after treatment, it may not be properly sealed so that bacteria can enter the broth from air and grow. Secondly, filtering requires great precision and unfiltered broth may enter the flask by chance.

The higher success rate by autoclaving from the class results suggests that autoclaving maybe a more reliable way of sterilisation. Autoclaving requires less human control, so imposes less human error.

Other ways to achieve sterilisation include heating in oven, using radiations (X-rays, gamma rays etc.) or adding antimicrobial agents (bactericidal, bacteriolytic) into the medium.


Question 1.

Liquid media used for culturing microorganisms in the lab is typically sterilized using a small autoclave (essentially, a programmable pressure cooker).

I prepare 5 litres of medium, split this into five one-litre bottles and put these into the autoclave. The autoclave is programmed for a 20 minute/121oC cycle.

My lazy colleague also makes 5 litres of the same medium, but instead puts the whole lot into a single large bottle and autoclaves this using the same programme.

A day-or-so after autoclaving, the medium in my colleague’s bottle has gone cloudy indicating that contaminating bacteria have survived the autoclaving and have grown in the medium. In contrast, the medium in all five of my bottles is clear (i.e. no bacterial growth).

Why do you think we have different results?

The 5-litre bottle is so large that heat transfer to the interior of the solution in retarded. As a result, the entire solution is not heated up to 121oC for sufficient time. By contrast, if splitting them into 1-litre bottles, the total surface area is larger such that the contents of the bottles are sufficiently heated to the required temperature within the 20-minute cycle.

What would you do differently if you wanted to autoclave a large volume (such as 5 litres) in a single bottle?

I would either programme the cycle of autoclave to be longer or set it to a higher temperature. This either allows more time for heating up the bottle, or the bottle can be heated up faster due to larger temperature difference.

Text Box: Question 2
In an imaginary experiment you carefully introduce a single cell of the bacterium Smartie red into a flask of sterile growth medium with the aim of producing a pure culture of this bacterium. Unfortunately, your lab partner sneezes at the critical moment and a single cell of Smartie green flies from his nose and into the flask. If S. red divides every 60 minutes and
S. green divides every 45 minutes, what is the ratio of S. red to S. green after 24 hours incubation (assuming unrestricted division of both bacteria throughout the 24 hours)?

In 24 hours, S. red divides 24 times while S. green divides 32 times. Hence, the ratio of S. red
to S. green after 24 hours incubation is 224 : 232 = 1 : 28 = 1 : 256

End of report.

Laboratory report: experiment 2 Reducing bacterial contamination

Text Box: Introduction
Briefly explain why it is important to use control measures to limit bacterial contamination of food; surfaces within the home; your hands, etc., and then outline the experimental strategy you used.

Bacteria are present all around us. It is virtually impossible for us to achieve sterility (i.e. absence of bacteria) as it is not attainable and practical, for example in fresh food. When sterility is not possible to achieve, we aim to achieve disinfection or decontamination where bacterial growth is limited. If bacterial growth is not limited, food may easily get spoiled and we are more prone to getting food-borne diseases and food poisoning. Bacterial growth can also help to spread diseases by other ways (e.g. air, water, direct contact) if the environment around us is abundant in bacteria.

In this experiment, some household treatments were tested for their potential antibacterial effects. Our group used Escherichia coli as the test microorganism. The following reagents were used – household bleach, TCP (an antiseptic), isopropanol, liquid soap, saturated salt solution, vinegar and honey. After overnight incubation, the presence of any clear zones and their sizes were noted. The presence of clear zones means absence of bacterial growth, hence effective inhibition of the reagent. As the reagent diffuses away from the centre, its concentration decreases radially – a larger clear zone means that the reagent works at lower concentration.
Text Box: Methodology
Briefly summarise the experimental method.

1.	8 nutrient agar plates were prepared. 50 µL of Escherichia coli bacteria suspension were spread across each plate using a sterile plastic spreader.
2.	The plates were left to dry for a few minutes, and then 10 µL of each of the following test solutions were added to the centre of a corresponding plate, water was used as a control.
3.	The plates were left to dry again, then the plates were incubated at 30°C for overnight, then stored at 4°C to stop further growth for a week.
4.	The diameter of zone of inhibition (“clear zones”) of each plate was measured (if any).
5.	Another set of data from a group using Micrococcus luteus was obtained.


Results Report your findings. What did the bacterial growth look like on the control (water only) plate? Which of the various treatments were effective in reducing the amount of bacteria? Was there a difference between the Gram positive M. luteus and the Gram negative E. coli?   The size of clear zones (diameter in mm) of different treatments were measured as follows:
11 mm6 mm12 mm6 mmnone6 mmnonenone
  The data of another group using M. luteus are as follows:
22 mm9 mm20 mmnonenone10 mmnonenone
  The class results are as follows (% of groups showing inhibition):
  Combining the three sets of results above – bleach, TCP, isopropanol are the most effective means of inhibiting bacterial growth (most groups have inhibition). They are all commercial products for disinfections. Soap & vinegar also show good inhibition. Salt solution & honey show some extent of inhibition, but are probably not effective means of bacterial control. All groups had bacterial growth on the control plate; the whole surface of the control plate appeared to be cloudy.   Comparing the two sets of data using E. coli & M. luteus, bleach & isopropanol are the most effective treatment for inhibiting growth of two species. TCP & vinegar also inhibit growth of these two species. Soap shows inhibitory effect on E. coli but not on M. luteus, although this maybe due to experimental variations only, so whether the treatments have differences on the two bacteria is inconclusive.
Text Box: Discussion
Discuss the significance of your findings in terms of the everyday use of the various chemicals /treatments.

The first three reagents – bleach, TCP and isopropanol are all commercial products for disinfections. As most groups show inhibition using these reagents, they are considered to be effective for bacterial control. Bleach and isopropanol are strong disinfectants, judging by their larger sizes of clear zones. TCP is a milder disinfectant.

Some household reagents are thought to have bacterial control effect but turns out they do not. For example, only some groups show inhibition of growth using soap, which it is always thought that it kills bacteria. Washing your hands with soap does not necessarily mean that they are clean.

Nevertheless, we cannot always use disinfects for bacterial control, for example in food preservation. In this experiment, although vinegar and salt solution do not show remarkable inhibition, we should not overlook their effects in bacterial control in food. When present in higher concentrations, vinegar and salt do really show some effects of preservation.
Text Box: Question:
Disinfectant products such as Dettol claim that it “kills 99.9% of bacteria”.
Discuss your interpretation of this claim, and how you opinion changes if it means “99.9% of all the different types of bacteria” or “99.9% of any bacterial species”.

From the first glance, it seems that it really means “killing 99.9% of all bacteria”. However, given the variety of bacteria is so diverse, I believe it is actually impossible to have a formula that could kill so many bacteria at room temperature and is still safe to be touched. I believe such claim is only based on selected bacterial species, perhaps easier to be killed. I would suggest the statement to be changed to “killing 99.9% of common houseshold bacteria”, to make it more sensible.

If the latter is true, and E. coli doubles its numbers every 30 minutes on the treated surface, how long before the effects of Dettol are nullified?

Suppose there were 1000 bacteria initially, after the treatment of Dettol, only 1 bacterium remained. The bacterium replicates itself in 30 minutes. As this goes on, there will be 2n bacteria after n×30 minutes. As 29 = 512 < 1000 and 210 = 1024 > 1000, the effects of Dettol are nullified after approximately 5 hours (10 divisions).

End of report.


Laboratory report: experiment 3

Colony growth and analysis of two bacterial strains

Text Box: Results
Report your findings. Which strain grew on which solid medium? How good was your streaking (were single colonies evident)? What was the appearance of the colonies (did all the colonies have the same morphology)? What were the results of the Gram strain and the Gregerson test? Any unexpected results?

Both strains grew well on the pure nutrient agar medium, with cloudiness covering around a quarter of the plate of M. luteus and three quarters of the plate of E. coli. Around 20 single colonies (each of diameter ~2 mm) are found on each plate. However, there are differential growths on the EMB plates – E. coli had well growth on the EMB plate, 15 single colonies were observed (diameter ~2 mm); M. luteus showed only limited growth on EMB plate, only around 20 small single colonies were observed (diameter <1 mm).

The colonies of M. luteus appeared to be more opaque and yellow in colour. By contrast, the colonies of E. coli appeared to be less opaque, white in colour and liquid- like.

Observing the chosen bacteria colonies after treatment with Gram stain, M. luteus appeared to be coccus and violet in colour, while E. coli appeared to be rod-shaped and pink in colour.

As for  the  Gregerson test,  after  treatment with  3%  KOH,  the mixture  with  E. coli
became sticky, while the mixture with M. luteus did not show any change in viscosity.
Text Box: Discussion
Discuss what your results tell you about the two strains – how do they compare? (colony morphology, cell morphology, etc.). Comment on the three different strategies for determining Gram type (i.e. Gram stain, Gregerson test, growth on EMBA).

The three tests we employed for determining Gram type all showed contrasting results for the two species, indicating they are of different Gram types. The results all point to E. coli being Gram negative and M. luteus being Gram positive. They differ in both colony morphology and cell morphology, as described above.

The three different strategies have different advantages & disadvantages. Gram stain is a reliable technique as the pink and purple colour can be easily distinguished under a light microscope. However, the staining procedure involves several steps, which is relatively time-consuming. Growth on EMB plates gives obvious results, which do not require further treatments, but the growth of bacteria maybe easily affected by other factors (e.g. contamination). Growth on EMB plates also requires incubation for several days. The Gregerson test is perhaps the easiest and fastest test, results can be obtained almost instantly. However, the degree of stickiness is subjective and maybe dependent on species. Among the three strategies used, I believe that Gram stain is the most reliable method for determining Gram type.


Text Box: Question 1.
Find out which of these bacterial species is Gram positive and which is Gram negative?

i)	Pseudomonas aeruginosa
ii)	Corynebacterium diphtheriae
iii)	Bacillus anthracis
iv)	Vibrio cholerae
v)	Rhodobacter sphaeroides
vi)	Anabaena sphaerica
vii)	Haemophilus influenzae
viii)	Staphylococcus aureus
(ii), (iii) and (viii) are Gram positive.
(i), (iv), (v), (vi) and (vii) are Gram negative.
Text Box: Question 2.
A standard Petri dish is 9 cm in diameter. If you poured a nutrient agar plate that was
0.5 cm thick, what volume of nutrient agar would be in the dish?

Volume of nutrient agar
= πr2h
= π(4.5)2(0.5)
= 31.8 cm3

What volume of a 50 mg/ml stock solution of antibiotic would you need to also add to the dish if you wanted a final concentration of 25 µg/ml?

amount of antibiotic required
= 25×31.8
= 795 µg
= 0.795 mg

volume of stock solution required
= 0.795÷50
= 0.0159 mL
= 15.9 µL

End of report.

Laboratory report: experiment 4 Isolation of Pseudomonas fluorescens

Text Box: Introduction
Briefly describe the basis of an enrichment medium and a diagnostic medium.  Outline the goals of the experiment.

An enrichment medium promotes the growth of a specific type of microorganism in interest. It usually contains a substance that is essential for growth of the specific type microorganism only. It restricts the growth of other microorganisms by the absence of essential nutrients or the presence of inhibitory substances.

A diagnostic medium is a growth medium that stimulates only the microorganism species in interest to produce an observable change, so that the presence of that type of microorganism can be confirmed.

This experiment aims to isolate the species Pseudomonas fluoescens by the principle that it is one of the few species that is able to utilise benzoate as carbon source. P. fluoescens was enriched in an enrichment medium with benzoate as the only carbon source. After proliferation in the enrichment medium, it was then transferred to a diagnostic medium to stimulate the production of green fluorescein pigment for identification. P. fluoescens was then characterised by Gram stain, catalase test and oxidase test. The results allow us to describe the colony & cell morphology of P. fluoescens, to state the Gram type, and to detect the presence of catalase & cytochrome c oxidase enzyme.
Text Box: Methodology
Briefly summarise what you did in the lab.

1.	Two sodium benzoate enrichment medium tubes were prepared. The medium tubes contain 0.2% sodium benzoate as carbon source, as well as other inorganic ions in a specific composition as stated in the coursebook.
2.	A grain of soil was added to one of the benzoate medium tubes.
3.	The two tubes of medium, one with soil & one without soil (as control) were incubated at 25°C for four days.
4.	The benzoate broths that showed growth (i.e. with cloudiness) were streaked onto King Ward Raney (KWR) agar.
5.	After three more days, the yellow/green pigmented colonies from the KWR agar were picked to carry out Gram stain, catalase test & oxidase test.


Text Box: Discussion
Discuss what your results tell you about the bacteria in the soil sample.

P. fluorescens is Gram-negative as shown by the results of Gram stain. The exceptionally well growth of P. fluorescens in the benzoate medium indicated that the bacterium is able to utilise benzoate, a non-common carbon source, for its own growth. This increases its competitiveness in a benzoate-rich environment.

P. fluorescens is also able to produce green fluorescein pigment as stimulated on the KWR plates.

The results of catalase and oxidase test suggest that P. fluorescens contains catalase and cytochrome c oxidase enzymes. The presence of cytochrome c oxidase suggests that P. fluorescens uses oxygen for energy production. Together with the presence of catalase, these suggest that P. fluorescens maybe an obligate aerobe.
Text Box: Results
Report your findings. Were you successful in obtaining growth in the benzoate medium? What was the morphology of the colonies and the cells? What was the appearance of the KWR plates? Results of the various tests?

There was bacterial growth shown in the benzoate medium with soil, while no bacterial growth shown in the control tube without soil. On the KWR plates, the colonies appeared to be yellowish-green. Around 10 single colonies were observed.

When viewed under microscope, individual cells appeared to be rod-shaped and pink after Gram stain, indicating that it is Gram negative.

P. fluorescens is positive towards catalase test. Bubbles were evolved when hydrogen peroxide was added to the cells.

P. fluorescens is also positive towards oxidase test. When TMPD was added to a test colony, purple colour was observed in around 30 seconds.

Question 1.

By visiting this webpage, answer the following questions about Pseudomonas fluorescens.

a)      Which explosive pollutant of soil could P. fluorescens degrade? Draw the structure of this compound.

P. flurorescens degrades TNT (trinitrotoluene) explosive. Its systematic name is 2-methyl-1,3,5-trinitrobenzene, and its structure is as follows:-

b)      Explain the two ways that P. fluorescens might act to suppress plant diseases.

P. flurorescens produces some secondary metabolites, including antibiotics, siderophores and cyanide, which protect them from fungal infection. P. flurorescens also colonises rapidly in soil, outcompeting other pathogenic species.

c)      How big (in Mb) is the genome of P. fluorescens strain PfO-1 (hint: click on ‘Info’)?

6.438405 Mb.

d)     If a typical protein-coding gene is ~1 kb, approximately how many genes would you expect P. fluorescens to have?

Approximately 6438 genes.

e)      Hospital infections of Pseudomonas species occur frequently and can occasional prove lethal (see: Approximately how many cases of infection are reported by the Health Protection Agency each year?

Around 3850 cases (between 3700 and 4000) are reported each year.

End of report.


Laboratory report: experiment 5 Isolation of Azotobacter species from soil

Text Box: Introduction
Nitrogen (N) is an essential macronutrient for all living organisms since it is a component of nucleic acids and proteins, and also many co-factors and secondary metabolites. The original source of nitrogen is the gas N2 that is the major component of the atmosphere, although most organisms acquire their nitrogen from uptake of N- containing organic compounds or by uptake of inorganic forms of N such as nitrate (NO32–), nitrite (NO22–), ammonium (NH4+) and urea ((NH2)2CO). A few groups of bacteria, including members of the genus Azotobacter, are capable of conversion of atmospheric N2 into ammonia (a process referred to as ‘nitrogen fixing’), and thus occupy a critical position in the N cycle of life.

Azotobacter are free-living soil bacteria and we can exploit their ability to fix nitrogen to develop a simple selection strategy for their isolation. Using a solid growth medium devoid of a source of N such as agar plates supplemented with Ashbey’s Nitrogen Free (ANF) medium, we can obtain actively growing Azotobacter colonies from a soil inoculum whilst limiting the growth of the many other bacterial groups within the soil sample.

In this experiment we successfully isolated Azotobacter colonies from soil and were able to determine their colony morphology, their cell morphology and their Gram status.
Text Box: Methodology

Soil grains were lightly sprinkled onto agar plates containing Ashbey’s Nitrogen Free medium and the plate incubated at 25oC for two days.

Large actively growing colonies were selected for Gram staining following the method described on page 96 of the BIOL1003 course booklet.

Text Box: Results
Report your findings. Were you successful in obtaining growth on the ANF plates? What was the morphology of the colonies? Result of the Gram stain and appearance of the cells?

There was successful growth of bacteria on ANF (Ashbey’s Nitrogen-free) plates, some large yellowish slimy colonies were formed. This indicated the colonies were able to grow by acquiring nitrogen from nitrogen gas in atmosphere (nitrogen-fixing).

Under microscope, Azotobacter appeared to be oval in shape under microscope, some of them are being spherical and rod-shaped. Their sizes were relatively large, about 1 to 2 micrometres in diameter. Applying Gram-stain, cells appear to be red in colour, suggesting Azotobacter is Gram-negative.
Text Box: Discussion
Discuss what your results tell you about the bacteria? Occasionally you see other bacteria growing in association with the Azotobacter colonies. Speculate why these other bacteria are able to grow on ANF medium even though they are not able to fix N2?

The success of growing on ANF plate (without nitrogen) suggested that bacteria
Azotobacter were nitrogen-fixing, and it is Gram-negative from our staining results.

Although on our colony we have chosen there appeared to be no other species growing in association with Azotobacter, there are still possibilities for other non- nitrogen-fixing bacteria to grow successfully on ANF plates. One of the possible reasons is that Azotobacter can secrete nitrogenous wastes, and some bacteria may take advantage of the nitrogenous wastes as their sources of nitrogen and grow around Azotobacter.


Text Box: Question 1.
Azotobacter species cannot form endospores, but have an alternative mechanism for surviving harsh conditions. Briefly explain this alternative mechanism.

Although Azotobacter species are not able to form endospores, they are able to form cysts which enable them to survive in harsh conditions. Cysts are known as the ‘resting structure’ of this species. Cysts have degree of similarities as bacterial endospores – they have negligible respiration rate and are resistant to desiccation, mechanical damage & radiation. However, cysts are not very heat-resistant.
Text Box: Question 2.
Visit the following website on nitrogen-fixing bacteria and then answer the questions below:

a)	Name three species of N2-fixing bacteria.

Azotobacter vinelandii, Klebsiella pneumonia, Rhizobium leguminosarum.

b)	What is the name for the N2-fixing enzyme?

Nitrogenase is the nitrogen-fixing enzyme.

c)	Which metal ion co-factors are found in three different forms of this enzyme?

The major form: iron and molybdenum The second form: iron and vanadium The third form: iron only

d)	Which gas causes the inactivation of the enzyme?

Dioxygen (O2) inactivates nitrogenases. O2 may oxidise the iron-sulphur cofactor of nitrogenases.

End of report

Laboratory report: experiment 6 Isolation of Bacillus from soil

Text Box: Introduction
Briefly describe the background and aims of the experiment.

Most soil bacteria are non-resistant to heat (typically at 80°C), as they are unable to form endospores. However, species such as Bacillus are able to form endospores, so are more resistant to heat. Endospores are in fact dormant structure of cells that are resistant to harsh environmental conditions. The difference between endospores and spores is that spores are offspring of a species while endospores are internal structures of bacteria. Bacillus endospores present in soil survive heat treatment can successfully germinate on nutrient agar plates.

The aim of this experiment is to isolate Bacillus endospores from soil and estimate the number of Bacillus endospores present in soil.
Text Box: Methodology
Briefly summarise what you did in the lab.

1.	One gram of soil (our sample was soil B)was added to 9 mL of Ringer solution. The salt components of Ringer solution can shock the bacteria osmotically. The mixture was placed in a 80°C water bath for 10 minutes.
2.	The heated test tube was cooled under tap water. 5 serial 10-fold dilutions were carried out.
3.	0.1 mL of samples from the last 4 dilutions were spread onto nutrient agar plates.
4.	The plates are incubated at 30°C for one week.
5.	After a week, colony numbers were counted. The number of spores per gram of soil sample was calculated.
6.	Bacillus endospores were tested using Gram stain, catalase test and spore stain.


Text Box: Discussion
Discuss how easy it was to obtain an accurate count and whether the various tests confirmed that the colonies were Bacillus species.

Plates of different dilution factors were prepared and only one of them was chosen for estimation of numbers. The plate that contained too many clustered colonies was not chosen because the actual number of colonies cannot be counted. The plates were too few colonies should not be chosen as well because this would give a greater error is estimation (e.g. difference between 1 and 2 colonies on a plate may give a two-fold difference in estimation). Therefore, in obtaining an accurate count, the suitable plate must be chosen. Also, another problem was that some colonies might stick together, which made us hard to differentiate if it was a single large colony or two small colonies. It is not easy to obtain an accurate count.

The morphology and the results of Gram-stain and catalase test matched that of Bacillus species. However these still cannot prove that is Bacillus species since these tests were not specific. The spore stain test gave us a good confirmation though, as red and green colour, which represented vegetative and endospores can be clearly differentiated under microscope. In the soil, the major species that produces endospores is Bacillus, so we could conclude the colonies were Bacillus species.
Text Box: Results
Report your findings. What was the appearance of the Bacillus colonies? What was your estimate for the number of spores in 1 g of soil? How do your results compare with the class results? What were the results of the spore stain, etc...?

On the plate with 2 serial dilutions (100-fold), there were large and clustered colonies. On the plate with 1000-fold dilution, there were 19 colonies. There were only 1 colony on the plate with 10000-fold dilution and no colony on the plate with 100000- fold dilution. Thus, the estimation of number of endospores was based on the plate with 1000-fold dilution.

The colonies were all in white colour, with a little elevation. When viewed under microscope, they are rod-shape cells. Applying Gram-stain, the cells became violet in colour, indicating Bacillus is Gram-positive. For the catalase test, bubbles were observed when hydrogen peroxide was added, indicating the species contains the enzyme catalase. Finally for the malachite green spore stain, red rod-shaped cells were obtained. Some cells had a green spot at the centre of the cell. Vegetative cells appeared to be red while endospores appeared to be green after spore stain.

The estimated number of endospores in 1 g of soil is 19×104 = 190,000 per gram. Compared with the class result, which was 306,000 per gram, our estimation was a bit lower. This may possibly due to incomplete transfer of soil into Ringer’s solution and error when carrying our serial dilution.

Question 1.

Using ‘Todar’s Online Textbook of Bacteriology’ (

Go to the contents page and then click on ‘Bacillus and related endospore-forming bacteria’ at the bottom of the page. Then answer the following:

a)      Explain why Gram staining of spore-forming species might prove problematic.

Endospores formed are strongly resistant to staining. Hence, endospores may appear as non-staining entities. This may affect the identification of properties of species.

b)      Mature endospores have no detectable metabolism. What is this state called?

This state is known as crytobiotic.

c)       List five differences between vegetative cells and endospores.

1)  Vegetative cells appear to be refractile under microscope while endospores appear to be non-refractile.

2)  Vegetative cells have high cytoplasmic water activity while that in endospores are low.

3)  There are enzymatic activities and macromolecular synthesis in vegetative cells while they are absent in endospores.

4)  Vegetative cells are generally sensitive to lysozyme actions, dyes and staining while endospores are resistant.

5)  Endospores have high resistance towards heat, acids and radiation while vegetative cells are susceptible to attack.

Text Box: Question 2.
One argument [the “Panspermia” (or more correctly “Exogenesis”) hypothesis] for  the how life began on Earth is that primitive life forms were transported here on meteorites that originated from another planet. Given what you now know about endospores, explain whether you think that this is a plausible hypothesis.
Endospores are known to have high resistance towards heat, desiccation and radiation. It is possible for endospores produced by species on other planets to be transferred to Earth by meteorites. Although when meteorites pass through the atmosphere, they encounter a high temperature and lots of radiation, endospores adhered on them are resistant from being attacked.

End of report


Laboratory report: experiment 7 Isolation of enterobacteria from river water

Text Box: Introduction
Provide a brief background to Enterobacteria; why they might be prevalent in river water, and their selection on MacConkey’s agar

Enterobacteria are a group of rod-shaped bacteria. They have Gram-negative stains and are facultative anaerobes. This group includes both pathogenic and symbiotic species, common examples are Escherichia coli, Salmonella and Shigella.

Enterobacteria are highly abundant in river water due to contamination, mainly by animal sewage. Other possible sources of pollution are agricultural runoff, discharge of human sewage into river etc.

The aim of this experiment is to isolate Enterobacteria from a sample of river water, and to determine the level of Enterobacteria contamination. River water samples were serially diluted, and grown on MacConkey agar plates. MacConkey agar plates are selective plates which allows only Gram-negative bacteria to grow. This helped us in isolating Enterobacteria. MacConkey plates also differentiate bacteria for lactose fermentation, if bacteria can utilise lactose, they will appear as red/pink, hazy colonies. Bacteria that cannot utilise lactose will utilise peptone on the plates instead, colonies will appear to be white colonies.
Text Box: Methodology
Briefly describe the method used to isolate the bacteria.

1.	A sample of river water was used (river water sample 1 for our group) to prepare 3 serial 10-fold dilutions using Ringer’s solution.
2.	0.1 mL in each of the dilution and the undiluted sample was plated out onto MacConkey agar plates.
3.	The plates are incubated at 37°C (human body temperature) for 2 days, then transferred to the refrigerator to stop growth and stored for 5 days.
4.	The colonies numbers were examined on each plate, the number of Enterobacteria per mL of river water was calculated using the plate with a countable number of colonies.


Report your findings. What was the appearance of the colonies on the plate? What percentage were lactose fermenters? What was your estimate for the number per ml of river water? How did this compare with the group results?

The number of colonies on the plate with undiluted river water were all clustered together and too many to be counted. Colonies in the plate after the first dilution were still too many to be counted. There were 28 distinct colonies on the plate with a 100- fold dilution; there were only 2 distinct colonies on the plate with a 1000-fold dilution.

Thus we consider the plate with 1000-fold dilution. The distribution of morphologies of the colonies is shown below.

 Number of coloniesPercentage
White colonies1036%
Red/pink colonies1864%

Lactose fermenters gave red/pink colour due to production of acid, while those bacteria that cannot ferment lactose produces ammonia using peptone, giving a white colour. For our sample, 64% of bacteria were lactose fermenting while the remaining 36% were not.

The estimated number of Enterobacteria in the river water sample is 28×103 = 28,000 per mL. Compared to the class results, which is 85,000 per mL, our results were lower. This may probably due to error in carrying out serial dilution.

Text Box: Discussion
Discuss your findings in the context of the appearance of the water sample and whether it would be advisable to drink it.

From our findings, there are far too many bacteria species in river water than we may have expected. Although sometimes river water appears to be ‘clean’ (highly transparent), they contain many Enterobacteria that cannot be seen with our naked eyes.

Enterobacteria such as E. coli, Salmonella are pathogenic species to humans. Safe- drinking water should be free of pathogenic bacteria or their concentration should be as low as possible. We should not judge if a water sample is safe to drink solely by its appearance, but we also have to know its source and whether it has undergone water treatment.


Text Box: Question 1.
The term ‘enterobacteria’ is based on the Greek word “enteron”. What does enteron mean?

‘Entero’ refers to intestine. Therefore ‘Enterobacteria’ means bacteria living in the intestine.

End of report

Laboratory report: experiment 8

Skin bacteria and their antibiotic sensitivity


Text Box: 7.	Some colonies on the plates with good sizes were examined under microscope; their morphology and Gram reaction were noted.
8.	One of the colonies chosen above was suspended into 1 mL of Ringer’s solution.
0.1 mL of this was inoculated on the sensitivity test agar.
9.	Different antibiotics discs containing 5 different antibiotics mentioned above were placed onto the agar after allowing the agar plate to dry for 15 minutes. The plate was incubated at 37°C for a week.
10.	The ‘clear-zones’ (if any) due to antibiotic resistance were noted and the diameters were measured.


Report your findings: the number and types of bacterial colonies observed before/after handwashing, etc.; the results of the analysis of the chosen colonies – Gram stain, morphology, etc. To which antibiotics were the bacteria resistant/sensitive?

The bacteria colonies found on the first sector that was pressed with unwashed finger were abundant. They were mostly white in colour; their sizes vary from <0.5 mm to 2 mm. There were only a few bacteria colonies found on the second sector that was pressed with washed finger. They were white in colour and their sizes were all less than 0.5 mm. On the third sector that was taken from the surface of a mobile phone, bacterial colonies were abundant. They were mostly red in colour, only one colony was white in colour. The sizes for the red colonies were <0.5 mm while for the white colony was 2 mm. Similar results were obtained from sector 4.

On viewing under microscope, the white colonies on sector 1 and 2 were spherical in shape and Gram-positive. The red colonies on sector 3 and 4 were rod-shaped and Gram-negative.

A white colony was chosen for the antibiotic test. The sizes of the clear zones are as follows.

Antibiotic usedQuantity usedSize of clear zone
Tetracycline10 µg14 mm
Penicillin1.5 units8 mm
Streptomycin25 µgNo clear zone
Rifampicin5 µg10 mm
Vancomycin30 µg10 mm

Therefore for the chosen colony, it was sensitive to all chosen antibiotics except for streptomycin.

Text Box: Discussion
Discuss your findings in the context of the possible health issues and treatment for infections arising from skin-borne bacteria.

As illustrated in the experiment, the bacteria found on our hands, mobile phone surface and banknotes (and other daily objects) are highly abundant and diverse. Therefore, we should be bare in mind that every object that we encounter daily may possibly have pathogenic bacteria on it. This normally does not do too much harm to us because our keratinised skin stops the entry of bacteria into our body. However, for infants and children, they may have a habit of thumb sucking which allows the entry of bacteria through their mouths. Parents have to take extra care on them, for example by cleaning the objects at home more frequently.

Bacteria on skin flora also pose a higher risk of infections in immunosuppressed people, for example AIDS patients or those whose skin was burnt. In treating this sort of infections, antibiotics are usually employed. However, doctors have to be aware of the issue of antibiotic resistance. In the sample colony chosen, the bacteria were resistant to streptomycin. The misuse of antibiotics may also help bacteria to develop antibiotic resistance, therefore doctors should advice patients with such infections to complete their treatment.

Question 1.

Antibiotics work by inhibiting a specific metabolic process found in bacteria cells.

Describe the mode of action of the following two antibiotics and briefly explain why human cells are unaffected.

1.      Penicillin

Penicillin is a type of β-lactam antibiotics, which inhibits cell wall synthesis. During cell wall synthesis, an enzyme called transpeptidase cross-links peptide chains. In the presence of penicillin, transpeptidase bind to penicillin and cannot continue to catalyse the cross-link reactions. As cell wall synthesis still continues, newly synthesised cell wall is not cross-linked and the strength is weakened. In addition, the binding of enzyme to penicillin also stimulates enzymes known as autolysins, which digest existing cell wall. Bacterial cells are eventually self- degraded. Human cells do not have cell walls, so are unaffected by penicillin.

2.      Streptomycin

Streptomycin is a type of aminoglycosides, which is a natural antibiotic produced by a bacteria called Streptomyces griseus. It inhibits protein synthesis, it binds to the 30S subunit of bacterial ribosome and interfere with tRNA binding to the 30Ssubunti. This results in misreading of codons, eventually inhibits protein synthesis and causes bacterial cell death. The structure of ribosomes in human cells (eukaryotes) are different from that of bacterial cells (prokaryotes), therefore human cells are unaffected by action of streptomycin.

End of report