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Reg. No:…………………………….
KIGALI INSTITUTE OF SCIENCE AND TECHNOLOGY
INSTITUT DES SCIENCES ET TECHNOLOGIE
Avenue de l'Armée, B.P. 3900 Kigali, Rwanda
INSTITUTE EXAMINATIONS – ACADEMIC YEAR 2013
END OF SEMESTER SUPPLEMENTARY EXAMINATION
FACULTY OF SCIENCE
SCIENCE-4-CHEMISTRY
THIRD YEAR, SEMESTER II
CHE 3329 BIOCHEMISTRY OF MICROORGANISMS
DATE:
/
/ 2013
TIME: 2 HOURS
MAXIMUM MARKS = 60
INSTRUCTIONS
1. This paper contains FOUR (4) questions and consists on TWO (2) sections
(SECTION A AND SECTION B).
2. Section A contains only one (1) question and it is a compulsory question.
3. Answer only two (2) out of the three (3) questions in section B.
4. No written material is allowed in exam room.
5. Write all answers in the booklet provided.
6. Do not forget to write your Registration Number.
7. Do not write any answers on this question paper
SECTION A
Question 1
(a) With reference to bacterial growth, distinguish between a growth factor and a growth
factor analogue
(4 marks)
(b) State three (3) differences between protozoa and other groups of microorganisms
(4 marks)
(c) Describe six (6) mechanisms of action of antimicrobial chemotherapeutic agents on each
the following groups of microorganisms;
(i) Bacteria
(6 marks)
(ii) Fungi
(6 marks)
SECTION B
Question 2
a)
State two (2) ways in which viruses differ from living cells.
(2 marks)
b)
Describe the functions of the viral capsid/envelope
(6 marks)
c)
State the characteristics/features that are used in classification of viruses (12 marks)
Question 3
(a) Describe four (4) factors affecting microbial growth and activity.
(b) A microbiological media has the following composition:
Constituent
Glucose
Ammonium phosphate monobasic (NH4H2PO4)
Sodium chloride (NaCl)
Magnessium sulphate (MgSO4.7H20)
Amount
5.0
0.5
5.0
0.2
Potassium phosphate dibasic (K2HPO4)
Water
1.0
1 liter
(10 marks)
Discuss the roles of each constituents listed in the above table on microbial growth and or
nutrition.
(10 marks)
Question 4
(a)
Describe the importance of staining microbial cells intended for microscopic examination
(2 marks)
(b)
Explain the steps involved in Gram-staining
(8 marks)
(c)
List and describe five (5) differences between the cell wall of gram positive and gram
negative bacteria.
(10 marks)
MARKING SCHEME
Question 1
(a) With reference to bacterial growth, distinguish between a growth factor and a
growth factor analogue
(4 marks)
- A growth factor is a naturally occurring compound (1 mark) that is capable of
stimulating bacteria growth (1 mark).
- A growth factor analog is a chemical agent that is related to (1 mark) and blocks the
uptake or utilization of a growth factor (1 mark).
(b) State three (3) differences between protozoa and other groups of microorganisms
(4 marks)
- Any of the following answers may be considered
- Protozoa are distinguished from prokaryotes by their eukaryotic nature (1 mark) and
usually greater size (1 mark).
- Protozoa are distinguished from algae by lack of chlorophyll (1 mark).
- Protozoa are distinguished from yeasts and other fungi by their motility (1 mark)
and absence of a cell wall (1 mark), and from slime molds by their inability to form
fruiting bodies (1 mark).
- Some protozoans have two nucleus; micronucleaus and macronucleaus (1 mark).
.
(c) Describe six (6) mechanisms of action of antimicrobial chemotherapeutic agents on
each the following groups of microorganisms;
(i) Bacteria
(6 marks)
- Inhibition of cell wall synthesis (1 mark).
- Inhibition of folic acid metabolism (1 mark).
- Destruction of cytoplasmic membrane (1 mark).
- Inhibition of RNA elongation (1 mark).
- Inhibition of DNA gyrase (1 mark).
- Interference with DNA-directed RNA polymerase (1 mark).
- Inhibition of protein synthesis (50S inhibitors) (1 mark).
- Inhibition of protein synthesis (30S inhibitors) (1 mark).
- Inhibition of protein synthesis (tRNA) (1 mark).
(ii) Fungi
(6 marks)
-
Inhibition of membrane functions; polyenes bind to ergosterol and disrupt membrane
integrity (1 mark).
Cell wall synthesis: Polyoxins inhibit chitin synthesis (1 mark).
Inhibition of egosterol synthesis (1 mark).
Disruption of microtube aggregation during mitosis (1 mark).
Inhibition of nucleic acid synthesis (1 mark).
Question 2
a)
Define two ways by which viruses differ from living cells (2 marks).
- Their simple, acellular organization (1 mark).
- The presence of either DNA or RNA, but not both in almost all virions (1 mark)
- In ability to reproduce independent cells and carry out cell division as prokaryotes
and eukaryotes do (1 mark).
b)
-
Describe the functions of the viral capsid/envelope (6 marks).
Protects nucleic acids from the effect of various enzymes, and chemicals (1 mark)
when the virus is outside the host cell (1 mark).
Helps to introduce the viral DNA or RNA (1 mark) into a suitable host cell (1
mark).
Stimulates the immune system to produce antibodies that can neutralize viruses (1
mark) and protects the host’s cells against future infections (1 mark).
c)
State the characteristics/features that are used in classification of viruses (12
marks).
(1 mark for each bulleted point)
- Nucleic acid characteristics. For example DNA or RNA viruses, single or double
stranded, molecular weight, segmentation and number of pieces of nucleic acid (RNA
viruses), the sense of the strand in ssRNA viruses.
- Capsid chemistry e.g. helical or icosahedral viruses
- The presence or absence of an envelope and ether sensitivity
- Their host. For example, animal, plant bacteria, insect or fungal viruses
- Diseases caused and/or special clinical features, method of transmission
- The viral size; diameter of the virion or nucleocapsid.
- The area of the host cell in which the virus multiplies.
- Number of capsomers in icosahedral viruses
- Immunologic properties
- Intracellular location of viral replication
- Presence or absence of a DNA intermediate (ssRNA viruses), and the
presence of
reverse transcriptase.
- Type of virus released.
Question 3
(a) Describe four (4) factors affecting microbial growth and activity.
-
(10 marks)
Temperature
(2.5 marks)
Need to state that for each microbial species or strain there is a minimum, optimum and
maximum growth temperature (0.5 mark).
Explanation on underlying factors leading reduced growth at minimum, optimum and
maximum temperatures required (1 mark).
Need to explain the different classes of microorganisms according to growth temperature;
pyschrotrophs, mesophiles, thermophiles and hyperthermophiles (1 mark).
-
-
-
pH
(2.5 marks)
Need to state that for each microbial species or strain there is a minimum, optimum and
maximum pH range for growth (0.5 mark).
Explanation on underlying factors leading reduced growth at minimum, optimum and
maximum pH ranges required (1 mark).
Need to explain the different classes of microorganisms according to pH. For instance
acidophiles, alkaliphiles and neutrophiles (1 mark)
Osmotic pressure
(2.5 marks)
Description of effect extracellar solute concentration on microbial growth required (0.5
mark).
Mechanism of osmotic effects on microbial growth required (1 mark)
Description of osmophiles and halophiles (1 marks)
Oxygen concentration
(2.5 marks)
Effect of oxygen concentration on aerobic, microaerophilic, aerotolerant, facultative
anaerobes, and obligate anaerobes (1 mark)
Detailed explanation on aerobiosis and anaerobiosis (1 mark) including toxic effects of
oxygen (1.5 marks).
(b) A microbiological media has the following composition:
-
Constituent
Glucose
Ammonium phosphate monobasic (NH4H2PO4)
Sodium chloride (NaCl)
Magnessium sulphate (MgSO4.7H20)
Amount
5.0
0.5
5.0
0.2
Potassium phosphate dibasic (K2HPO4)
Water
1.0
1 liter
Discuss the roles of each constituents listed in the above table on microbial growth and or
nutrition.
(10 marks)
Two marks for each bullet points
Ammonium phosphate acts as a nitrogen source required for the synthesis of amino acids.
The phosphate may also provide some buffering action.
The role of sodium chloride for osmotic control, to prevent osmotic lysis.
Magnesium sulphate acts as a source sulphur for synthesis of sulphur containing amino
acids such as cysteine
Potassium phosphate dibasic act as a buffer
Water helps in dissolution of components and helps in hydrolytic solutions
Question 4
(a)
Describe the importance of staining microbial cells intended for microscopic
examination
(2 marks)
- The microbial cell is translucent leading to poor visibility under the light microscope (1
mark). However strains of microbial cells leads to clear visualization under the light
microscopic examination (1 mark).
(b)
-
(c)
-
Explain the steps involved in Gram-staining
(8 marks)
Fixation of the microbial cells on the microscope slide (0.5 marks)
Addition of the primary dye usually crystal violet (0.5 marks)
After short time (30 seconds to 1 minute) the purple dye is washed off (0.5 marks).
The smear is covered with iodine (0.5 marks) a mordant which helps in fixation of the
dye (0.5 marks).
Iodine is washed off (0.5 marks) and both gram-positive and gram-negative will appear
dark violet or purple (0.5 marks)
Next the smear is washed off using alcohol or alcohol-acetone solution (0.5 marks)
This called the decolorizing agent (0.5 marks).
In gram-negative the crystal violet-iodine complex is washed off by the alcohol while in
gram positive it’s not washed off (0.5 marks).
This means that the gram-negative cannot be viewed under the microscope (0.5 marks)
and therefore there is need to apply a counter stain which is usually safranin (0.5 marks).
The alcohol is rinsed off, and the slide is then stained with safranin which is a basic dye
(0.5 marks).
The smear is the washed off blotted and dried (0.5 marks).
After microscopic examination, gram positive will appear blue (0.5 marks) while gram
negative will appear red (0.5 marks).
List and describe five (5) differences between the cell wall of gram positive and
gram negative bacteria.
(10 marks)
1 mark for each bulleted point
In gram-positive the peptideglycan layer is thin and multi layered while in gram-negative
its thin
Teichoic acids are present in the gram-positive bacteria but absent in gram-negative
bacteria.
Periplasmic space is absent in gram-positive but present in gram-negative bacteria.
Outer membrane is absent in gram-postive bacteria but is present in gram-negative
bacteria.
Lipopolysaccharide is virtually non-existent in the cell wall of gram-postive bacteria but
is present in the cell wall of gram-negative bacteria.
Two flagella rings are present in the basal body in gram-positive while four flagella rings
are present in the basal rings of gram-negative bacteria.
Gram-positive bacteria produce only exotoxins while gram-negative bacteria produce
both exotoxins and endotoxins.
Gram-positive bacteria are highly resistant to physical disruption while gram negative
bacteria are not resistant to physical disruption.
-
Cell walls of gram-positive bacteria are highly susceptible to disruption by lysozyme
while gram-negative bacteria are less susceptible to disruption by lysosyme.
Cell walls of gram-positive bacteria are highly susceptible to penicillin and sulfonalmide
while gram-negative bacteria are less susceptible to penicillin and sulfonalmide.
Gram-positive bacteria are less susceptible streptomycin, tetracycline and
chloramphenical while gram-positive bacteria are more susceptible.