Biol3400_midterm040310as

Biology 3400 Midterm
March 4, 2010
Time: 75 minutes Instructions 1. Do not open the exam booklet until you are instructed to do so. 2. Put your name, signature and student number on the exam booklet cover page, the last page, the start of the written answer section and the multiple choice answer sheet. Print your name on the last page of the exam booklet. 3. This examination is out of a total of 65 marks and consists of two parts:
4. Mark your multiple-choice answers on the multiple choice answer sheet.
5. Respond to the written answer questions as completely as possible on the exam booklet in the
space provided.
6. Your exam booklet contains 6 pages with printing on both sides of the page. Hand in the
entire exam booklet and the multiple choice answer sheet.
7. Carefully read and answer the questions as completely as possible and to the best of your
abilities.
PART I
Multiple Choice - 24 questions (1 Mark/Question)
Choose the BEST answer to each question. Mark this answer on the accompanying multiple
choice answer sheet.
Refer to questions posted outside the Microbiology Laboratories
Biology 3400 Midterm Written Answer Questions
March 4, 2010

PART II: Written Answer Questions. Answer as completely as possible in the space
provided below each question.

Question 1: Fill in the blanks with the best possible answer.
(6.5 Marks - 1/2 mark per blank)
Scientific name of a Gram-positive bacterium Researcher credited with discovering penicillin An antibacterial compound that inhibits DNA Sporosarcina, Oscillospira, Desulfotomaculum The general name for a medium to which the exact chemical composition is known A type of rRNA found in the small subunit of a prokaryotic ribosome Antibiotic that blocks protein synthesis at the transcriptional level The names of two fermentation pathways that produce lactic acid as a fermentation product homolactic (homofermentative), mixed acid The two major components common to all viruses Question 2:
a) Carbon monoxide (CO) is a potent inhibitor of electron flow and thus the establishment of
proton gradients in cells. It is extremely toxic to obligate aerobic forms of life and nontoxic to
obligate fermenters. Explain why CO affects these two different groups of organisms (i.e.,
obligate aerobe and obligate fermenter) in different ways. Be as specific as possible in your
explanation.
(6 Marks)

Carbon monoxide is very toxic to obligate aerobes because these organisms rely exclusively on
aerobic respiration for the generation of ATP (1 Mark). This means that these organisms use the
electron transport chain and oxidative phosphorylation to produce the bulk of their ATP (1
Mark). CO will disrupt oxidative phosphorylation/chemiosmosis by preventing the formation of
a proton gradient (1 Mark). This will ultimately shut down the production of ATP in these
organisms.
Carbon monoxide is not toxic to obligate fermenters because these organisms rely solely on
fermentation and substrate level phosphorylation to produce ATP (1.5 Marks). Since these
processes do not require proton gradients for ATP production, CO is not toxic to these organisms
(1.5 Marks).
b) Carbon monoxide also inhibits the movement of the bacterial flagellum. Explain the specific
mechanism behind this effect.
(2 Marks)
The energy required to drive the basal apparatus and rotation of a bacterial flagellum comes from
a proton gradient across the plasma membrane (i.e., proton motive force) (1 Mark). If CO
disrupts the establishment of proton gradients in cells then it will also interfere with the
movement (i.e., rotation) of a bacterial flagellum (1 Mark).
Question 3:
a) You are working with a bacterium that has minimum, maximum and optimum growth
temperatures of 10°C, 42°C and 30°C, respectively. Compare the growth of this bacterium at 30°C,
15°C and 45°C over time by drawing representative batch growth curves at each temperature. Use
optical density as the measure of growth and start each growth curve at the tick on the Y axis.
(7.5 Marks)
Question 3 (continued)
b) How would diluting the growth medium 10 fold with water (i.e., 1 part medium: 9 parts water)
affect the shape of the growth curve at 30°C?
(2 Marks)

Nutrient concentration has an effect on both rate of growth and yield (Fig 6.14). One should
expect to see a lower rate of exponential growth (1.5 mark) and a lower maximum OD achieved
(i.e., yield; 1.5 mark). The lag phase could also be extended as the cells adjust their metabolism
rate in response to the reduced nutrient concentration in the diluted medium.

Question 4
a) What are the similarities and differences between the cell walls of a Gram positive bacterium
and a fungal cell. Be sure to include information on structure, composition and function.
(6 Marks)

There are a number of points that can be made regarding similarities and differences between
these structures. The following represents some of the possible answers to this question but is in
no way an exhaustive list. A maximum of 4 marks may be earned for either similarities or
differences.
These structures are similar in the following ways (1 Mark/similarity):
• Location – they are both located outside the cell membrane
• Function – they both serve a protective role preventing the cell from mechanical and osmotic
• They are both composed partially of polysaccharides that contain sugar monomers bonded by β1-4 linkages (1 Mark) and one of the monomers is N-acetylglucosamine (1 Mark)

These structures are different in the following ways (1 Mark/similarity):
• The bacterial cell wall is composed predominantly of peptidoglycan whereas the fungal cell
• Peptidoglycan is composed of a heteropolymer consisting of N-acetylglucosamine and N- acetylmuramic acid bonded by β1-4 linkages whereas chitin consists of N-acetylglucosamine monomers bonded by β1-4 linkages. • Peptidoglycan is further reinforced by tetrapeptide cross bridges between N-acetylmuramic • Techoic acid is a major component of the cell wall of a Gram-positive bacterium • The cell wall of a Gram-positive bacterium is sensitive to β-lactam antibiotics whereas the Question 5: You are hired as a microbiologist and are working on a project to develop new
silage inoculants. Your assistant has given you a new liquid culture that has the following
characteristics. The cells from the culture tested positive for the presence of N-acetylmuramic
acid, the lack of phosphofructokinase activity, ethanol, lipopolysaccharide, calcium dipicolinate
and the presence of 18S rRNA sequences.
a) What does each element below tell you regarding the cells in the culture?
(6 Marks – 1.0 marks for each answer)
N-acetylmuramic acid – One of the modified sugars monomers found in the carbohydrate component of peptidoglycan. Only found in bacteria with cell walls the lack of phosphofructokinase activity – Phosphofructokinase is a key enzyme of the Embden Meyerhof glycolytic pathway. The cells in this culture are lacking in this glycolytic pathway ethanol – some cells are capable of producing ethanol, possibly by the ethanolic, heterolactic or mixed acid fermentation pathway lipopolysaccharide – a component of the outer membrane of Gram-negative cells calcium dipicolinate – A common chemical found in endospores the presence of 18S rRNA sequences – a species of rRNA found in eukaryotic organisms
b) Is the culture a pure culture? Why or why not? Be specific and briefly explain how you
arrived at your answer!
(2 Marks)

The presence of a number of factors strongly suggests that the culture is not pure. There is
evidence that the culture contains bacteria (N-acetylmuramic acid is only found in bacteria) as
well as eukaryotic cells (18S rRNA is found only in eukaryotic cells) (1 Mark). There is also
evidence that the culture contains Gram-positive (typically found in Gram-positive cell walls)
and Gram-negative bacteria (LPS is usually found in the outer membrane of Gram-negative
cells) (1 Mark)
Question 5 (continued)
c) Briefly describe the simplest way that you could test your conclusion about the purity of the
culture. Explain your answer.
(3 Marks)

The simplest method of testing the above conclusion is to prepare a smear of the culture and
Gram stain it (1.5 Marks). If the culture is a mixed culture of Gram negative and Gram positive
cells this will be very evident upon examining the Gram stain as both Gram positive and Gram
negative cells will be evident in the stained smear (1.5 Marks).
Alternatively one could also do a streak plate and look at the resulting colonial morphologies
(1.5 Marks). If the culture is a mixed culture then one will likely observe multiple colonial
morphologies on the plate (1.5 Marks).


Multiple Choice (Max = 24)
Written Answer Total (Max = 41)
Examination Total (Max = 65)

Source: http://people.uleth.ca/~selibl/Biol3200/ExamQues/MidTerm10AS.pdf