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nordsb [41]
3 years ago
8

Please help!!!!!!!! Will mark brianlist ASAP

Biology
2 answers:
Elanso [62]3 years ago
6 0

Answer:

the answer is D.........

KiRa [710]3 years ago
3 0

I'm going with C. It may be wrong but I think it's C.

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1. You are given the task to quantify the bacteria in your soil sample. You do a 10-fold serial dilution of your soil sample in
Irina18 [472]

Answer:

A serial dilution and the following plating experiment is done in order to determine the actual amount of bacteria/microbes in a specific volume of soil sample. As a standard test, a set volume of the lower dilution is obtained and positioned on a median plate and permitted to increase for the needed amount of time. The amount of colonies is evaluated and thus the overall Colony Forming Units (CFU) is determined by unit volume of the sample plated and therefore the volume of soil sample employed.

Therefore,

The calculation is done using:

CFU/ml = Number of colonies appeared × dilution factor / volume plated

Given,

Number of colonies appeared = 97

dilution factor = 10^(-6)

volume plated = 1/10 = 0.1

This will help us calculate it as:

CFU/ml = 97 * 10^(-6) / 0.1

= 97 * 10^(-7) CFU/ml

This CFU/ml helps determine the amount of bacterial colonies per unit volume of sample plated.

Given,

The original sample as 1g or 1000 mg of soil in its total volume.

Therefore,

The number of bacteria can be calculated using:

Amount of bacteria in original sample

= 97 * 10^(-7) CFU/ml × 1/ 1000 mg

= 9.7 * 10^(-3) CFU/mg

8 0
3 years ago
4. Does a prokaryotic cell have a cell wall?
vladimir2022 [97]
They have a cell wall and nucleus but they have no membrane-bound organelles like eukaryotic cells
6 0
3 years ago
What is autoregulation of blood flow?
earnstyle [38]
A manifestation of local blood flow regulation.
3 0
4 years ago
Hemophilia and colorblindness are both X-linked, recessive traits. A hemophiliac woman marries a colorblind man. Assume that bot
Gre4nikov [31]

Answer:

Normal daughters and Hemophiliac son

Explanation:

Females exhibit hemophilia and colorblindness in homozygous genotype only.  

The genotype of hemophiliac woman= X^hX^h

The genotype of colorblind man= X^cY

A cross between X^hX^h and X^cY would produce progeny in the following phenotype ratio= 1 Normal but carrier daughters:1 Hemophiliac son

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Stem cells may prove useful in the treatment of Type 1 diabetes because
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Because they can be differentiated to form insulin-producing cells
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