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leonid [27]
3 years ago
6

You are testing a new antibiotic in the microbiology lab. After treating Salmonella cells with the antibiotic, you wanted to see

how many viable cells remain. Which of the following methods will provide the most accurate count of viable cells?
A. Fluorescence-activated cell sorting of unstained cells.B. Serial dilution followed by an optical density measurement.C. Counting cells directly using a hemocytometer.D. Serial dilution followed by a colony forming units calculation.
Biology
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

D. Serial dilution followed by a colony forming units calculation.

Explanation:

The best way to count bacterial cells such as Salmonella accurately is through serial dilution and calculation of colony forming units.

Serial dilution is made using a petri dish containing appropriate culture medium for Salmonella growth. In this petri dish, with the help of an inoculation loop, the suspension containing the Salmonella will be scratched in the middle of the petri dish. Then, the inoculation loop will be passed over the salmonella streaks and will be dragged to the other side of the plate, where the loop will be scratched halfway through this space. The inoculation loop should again pass over the new salmonella streaks and be dragged through the rest of the petri dish. Then one must wait for the bacterial colonies to grow inside the plate.

Once these colonies are grown, those that appear in isolation will be counted through a specific calculation for the count of salmonella colonies.

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Neural crest and peripheral nervous system.

Explanation:

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8 0
3 years ago
Bearing fruit benefits plants because
Pavlova-9 [17]

Answer:

There are many benefits of bearing fruits. Please read the explanation section.

Explanation:

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4 0
3 years ago
Sickle cell anaemia is an inherited autosomal recessive
alekssr [168]

Answer:

1/8 (12.5%)

Explanation:

An autosomal recessive disease is an inherited disease in which an individual need to receive both defective alleles at the same gene <em>locus</em> to be expressed in the phenotype. In this case, both parents are carriers of the recessive mutant allele associated with the sickle cell anaemia trait, thereby both parents are heterozygous, ie., each parent has one copy of the normal allele 'H' and one copy of the defective mutant allele 'h' associated with this condition. In consequence, their first child has a 1/4 (25%) chance of having sickle-cell anaemia. Moreover, the chance of having a girl is 1/2 and the chance of having a boy is 1/2, thereby the final chance of having a girl sickle cell anaemia individual is 1/4 x 1/2 = 1/8 (12.5%).

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- F1 = 1/4 HH (normal); 1/2 Hh (normal); 1/4 hh (sickle cell anaemia) >>  

- Sex proportion of sickle cell anaemia individuals =  1/8 female sickle cell anaemia individuals + 1/8 male sickle cell anaemia individuals (1/8 + 1/8 = 1/4)

3 0
3 years ago
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