2. You have a stock solution of 1 x 10\' spores/ml. You serially dilute the spor

Question

2. You have a stock solution of 1 x 10' spores/ml. You serially dilute the spores as described in the table below Dilution 1 |:10 Dilution 3 Dilution 2 Dilution ID 100 1 dilution 3 100 1 stock solution | 900 1 water | 100 1 dilution 1 100 1 dilution 2 + 900 1 water | 900 1 water 900 1 water + + 2A. (2 pts) You load 10 1 of the stock solution onto a hemocytometer. How many spores do you expect to see in a 4 nl volume? 1000 PNG 2B. (2 pts) You plate 100 1 of dilution 4 on a YPD plate. How many colonies do you expect to see after incubation? 1000 sVO 1000 nl

Explanation / Answer

2a)

1 ml = 1 x 107 spores

1 ml = 106 nl

106 nl (1ml) contains 1 x 107 spores, the number of spores present in 4 nl is calculated as follows,

106 nl = 1 x 107 spores

4 nl = ? spores

(4 nl x 1 x 107 spores) / 106 nl = 40 spores.

2b)

1 ml = 1 x 107 spores

1 ml = 1000 ul

1000 ul (1ml) of stock solution contains 1 x 107 spores and from this, 100 ul is taken to make 1st dilution.

1000 ul = 1 x 107 spores

100 ul = ? spores

(100 ul x 1 x 107 spores) / 1000 ul = 106 spores

Therefore, 100 ul contains 106 spores and to which 900 ml water is added to make dilution 1.

Dilution 1 - 1:10 = 106 spores.

Similarly, 100 ul of dilution 1 is taken and 900 ml water is added to make dilution 2.

Dilution 2 - 1 : 100 = 105 spores.

Dilution 3 - 1:1000 = 104 spores.

Dilution 4 - 1:10,000 = 103 spores.

100 ul of dilution 4 contains 102 spores.

Colonies forming units (CFU) = number of colonies (n) / volume transferred to plate (s) x dilution blank factor (d)

CFU, concentration = n / s x d

n = CFU x s x d = 102 x 100 ul x 1/10,000 = 1

Therefore, number of colonies is 1.

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