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12 Cards in this Set
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Fungal growth
(E) . How are the following determined/calculated? a) The hyphal growth unit (G) b) The colony radial growth rate (Kr) |
(E) equals E = Gµ
The mean rate of hyphal extension (E) is the product of the hyphal growth unit (G) and the specific growth rate (µ). |
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The hyphal growth unit (G) ?
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G = Total length of the mycelium/number of branches
G=hyphal growth unit |
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(µ).
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µ=specific growth rate (h-1)
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U equals 0.201 and G equals 316 What is E?
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E equals GU equals 316x0.201 equals 63.5
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Width of the peripheral growth zone, w
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w = The length of hypha at the colony margin that that supports maximum growth
It is used to determine colony radial growth rate, Kr i.e |
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What does Kr equal?
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Kr = wu
u=specific growth rate |
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The data from table 6 was used to determine specific growth rate which was found to be 0.75hr-1
Determine the culture generation (τ) and cell generation time (td). |
τ = 1/0.75
td = τ x 0.693 = 1.33 x 0.693 = 0.924h. |
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The data in table one below were recorded in an experiment to determine the mass of a 10 cm3 suspension of Escherichia coli. Showing your workings at each stage, calculate the dry weight of the suspension in mg per cm3.
Weight (g) Tube 75.135 Tube plus dry cells 75.174 |
Tube plus dry cells – tube = 75.174 – 75.135 = 0.039 g = 39 mg
10 cm3 of suspension has a dry weight of 39 mg hence dry weight = 3.9 mg per cm3 |
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Table four below shows the dry weight obtained when Escherichia
coli was grown in a series of batch cultures in defined medium containing a range of concentrations of glucose. Complete the table to show the cell yield at each glucose concentration. Glucose (mg cm-3) 0.5 1.0 2.0 3.0 Dry weight (mg cm-3) 0.18 0.36 0.70 0.80 Cell yield (mg cells produced per mg glucose consumed) ??? |
Glucose (mg cm-3)
0.5 1.0 2.0 3.0 Dry weight (mg cm-3) 0.18 0.36 0.70 0.80 Cell yield (mg cells produced per mg glucose consumed) 0.18/0.5 = 0.36 0.36/1 = 0.36 0.7/2.0 = 0.35 0.8/3 = 0.26 |
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Explaining your reasoning, state whether glucose is the growth limiting substrate at this range of concentrations
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As the yield is constant between 0.5 and 2.0 mg cm-3, glucose is the growth limiting substrate at these concentrations. However, the direct proportionality between glucose concentration and cell yield no longer applies at 3.0 mg cm-3 indicating that glucose is no longer the growth limiting substrate at this higher concentration.
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Escherichia coli was grown in a continuous culture vessel where the vessel volume was 100 litres. The carbon source, glucose was 3.1g per l.
Determine the dilution rates when the flow rate of incoming medium was adjusted to 10, 20 and 40 litres per hour. Batch cultures had shown that 1 litre of this medium gave 1.12g dry weight of bacteria, umax was 0.85 per hour and ks was 0.012g per l. Determine the steady state bacterial concentration at each dilution rate. What were the mean residence times? What was the output? What was the critical dilution rate? |
Answer:
Hint. What was the Yield? 1.12/3.1 = 0.36 Dilution rates: D = F/ V, 10/100 = 0.1, 20/100 = 0.2, 40/100 = 0.4 per hr Mean residence time = 1/D For D = 0.1 , 1/0.1 = 10h, For D = 0.2, 1/0.2 = 5h For D = 0.4, 1/0.4 = 2.5h x = Y [Sr - ks (D/umax - D)] x = 0.36 [3 - (0.012x0.1/0.85 - 0.1)] x = 0.36[3 - 0.0012/0.75] x = 0.36[3 - 0.0016] x = 0.36 [2.9984] x = 1.079 The rate of production of biomass is called output and can be determined from DXx in the example D = 0.1, x 1.079, hence output Dx = 0.1 x 1.079. The critical dilution rate can be calculated from Dc = umax (Sr/ks+Sr) Dc = 0.85(3/0.012 + 3) Dc = 0.85x3/3.012 Dc = 2.55/3.012 Dc = 0.847 |
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In a continuous culture, the net change in substrate concentration is defined by?
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i. concentration of incoming substrate
(input) ii. minus the substrate concentration lost Net change in substrate concentration = input –output – consumption |