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Criado por sophietevans
mais de 11 anos atrás
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Questão | Responda |
Why is a growth curve produced rather than a stepwise graph? | Because the cells do not all DIVIDE at the same time, even though they are growing continuously. |
In the stationary phase, what happens to the number of viable cells? | It remains constant. |
What are some reasons for the stationary phase? | The population may be dividing and dying equally, the population may have ceased to divide but remains metabolically active, there is nutrient limitation (most frequently O2), there is accumulation of toxic waste products. |
Like the exponential growth rate, the death rate is also usually... | ...logarithmic. A constant proportion of cells dies every hour. |
Why might a small number of cells remain alive/the death rate slow towards after a significant period of death? | Cell turnover as lysed cell contents provide a nutrient source, or resistance. |
What is growth defined as? | An increase in cellular constituents. This may result in an increase in a microorganism's size, population number, or both. |
Cell multiplication is a consequence of... | ...growth. |
There is geometric progression in the generation time because... | ...the cell division time is constant. |
What happens in binary fission? | The bacterial cell elongates, the chromosome is replicated, two chromosomes separate, a septum forms in the mid-cell region and the parent cell divides into two identical daughter cells. |
What is a Z ring? | A constricting arrangement of PhZ proteins which constrict to split the one large cell into two after chromosomal replication. |
What is the calculation for the number of generations? | Number of generations (N) = (log(population after period of growth) - (log (initial population))/log2 |
How does the growth of the peptidoglycan cell wall in a growing cell vary between bacilli and cocci? | In bacilli, the growth is in discontinuous sections and eventually the evagination happens in the middle. In cocci, half of each of the daughter cells produced will be of the original cell and half will be newly synthesised. |
What is the division time of E coli? | ~20 minutes. |
What is the calculation for the number of populations after a period of growth? | logNt = (initial population number x 2 ^ the number of generations in the growth period) |
What is the division time in M. tuberculosis? | ~24 hours. |
If chromosomal replication takes 40 minutes in E coli, how is its division time around half that? | Continuous overlapping chromosomal replication occurs. |
In a closed system, why is population growth not exponential? | The exponential growth phase does not last for long due to the accumulation of wastes and because of nutrient depletion. |
The rate of growth in a batch culture can be expressed in terms of the mean growth rate constant (k) i.e. the number of generations per unit time. What is the calculation for this? | k = (log (number after period of growth) - log (initial number) / (log2 x (time between initial population number and number being measured) k = (logNt - logN0)/log2xt |
Which two characteristics can be used to measure population growth? | Mass or number. |
What is the cell number? | The number of viable cells per unit volume of culture. |
What is the calculation for estimating population size by counting prokaryotes in a Petroff-Hausser chamber and counting eukaryotes in a haemocytometer? | (average number of cells per square) x (volume of haemocytometer square) x dilution factor Example: (224/80) x (1/(2.5 x 10-7)) = 11200000 = 1.12 x 10-7 |
What are two problems with haemocytometer counts for estimating the size of a population? | Besides the experimental error that can be associated with counting chamber use, the microbial population must be fairly large for accuracy as the dilutions decrease the number greatly, and it is difficult to distinguish between viable and dead cells. |
Using a spectrophotometer to estimate microbial population size by turbidity is inaccurate because... | ...turbidity can vary between the growth and death of a culture and death can occur without affecting turbidity. |
Which 4 methods are common for performing a viable count? | Spread plate, pour plate, Miles & Misra plate, spiral plater. |
Why must cultures be serially diluted before counting? | Because the original sample will inevitably contain so many organisms that confluence will prevent accurate counting. 0.1ml is used on a spread plate, 1ml is combined with 9ml agar in a pour plate and 0.02ml is collected in an inoculating loop for Miles & Misra. |
Why is a physiological saline solution preferred for dilutions? | It is not harmful, it exerts a normal osmotic effect but does not encourage growth (unlike a nutrient broth) and so the cell count will accurately reflect the original number after dilutions. |
When determining the total from a plate count, it is assume that 1 colony arose from... | 1 cell. |
Units of VIABLE counts are expressed as? | Colony forming units (CFU)/mL |
If you have the rate, how do you calculate the time for 1 generation to develop/double? | G is the inverse of K, so G = 1/K. |
List some inaccuracies of performing counts. | Cells may be clumped together rather than well dispersed, one cannot be certain that each colony arose from an individual cell, low counts may arise if the agar medium cannot support the viable microorganisms present, and hot agar may injure/kill sensitive cells in pour plates. |
What else can be measured to measure population growth, other than mass/turbidity/colonies? | Quantification of protein, DNA or ATP, or of chlorophyll in photosynthetic protists. This is a long a complex process of washing and centrifuging which is relatively sensitive but difficult and time-consuming. |
What are the four stages of bacterial growth in batch culture? | Leg, exponential growth, stationary phase and death. |
List some environmental conditions that may affect microbial growth? | Water availability, pH, temperature, oxygen concentration, pressure, radiation. (Nutrient supply and waste removal also) |
Why might the lag phase occur? | The cells may be old and depleted of ATP/essential cofactors/ribosomes which need to be synthesised before growth can begin, there may be a change in nutrient medium which requires a different enzyme to be produced or a mutation for its production may need to occur, the microorganisms may have been injured and require time to recover, the microorganisms might have been dormant (e.g. refrigerated) and may need time to adjust, the inoculant may be in spore form and require time to germinate before growth. |
During the exponential/log phase, the microorganisms are dividing at the maximal possible rate, given which factors? | Their genetic potential, the nature of the medium, and the conditions under which they are growing. |
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