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ExtremeBDS [4]
2 years ago
6

A virologist has a cell-free extract of herpes simplex virus (HSV) that has a titer of 6 x 106 infectious virus particles/mL. Th

is virus forms plaques on monolayers of the monkey kidney cell line, CV-1. Following completion of a 10-fold dilution series, the virologist plates 100 microliters of each sample in duplicate.
Required:
What experimental results did he MOST LIKELY observe in the duplicate plates corresponding to his 1:10000 dilution?
Biology
1 answer:
CaHeK987 [17]2 years ago
3 0

Answer:

61 for Plate 1 ,

59 for Plate 2

Explanation:

Given that the original stock has a concentration of  6 * 10^6 infectious virus particles /mL

and it was diluted by 1:10000

New concentration = ( 6 * 10^6 ) / 10000 = 600 infectious virus particles/ mL

Also given that the virologist plates 100 mL of each sample

therefore; 0.1 mL has a count of 60.

The experimental results that he is most likely to observe in the duplicate plates will be ;

61 for (Plate 1) ,

59 for (Plate 2)

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Answer:

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Question: What will be their children’s possible phenotypes?

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Here H, indicate normal gene and small h indicate defected (hemophilia) gene.

Parents:                      XHXh                   X           XhY

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Phenotype of offspring:

XHXh:  The child will be daughter and normal (25 % chances)

XHY:     The child will be son and normal (25 % chances)

XhXh:   The child will be girl and Hemophiliac (25 % chances)

XhY:      The child will be son and Hemophiliac (25 % chances)

Hope it helps!

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