You do know that you have to give a topic on what this project is about, also doing this quick is never gonna happen. Science projects take TIME AND EFFORT. So, with that being said, sorry for wasting your time.
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
A temperature inversion is a thin layer of the atmosphere where the normal decrease in temperature with height switches to the temperature increasing with height. An inversion acts like a lid, keeping normal convective overturning of the atmosphere from penetrating through the inversion.
Answer:
A. VG = 80
B. Broad sense heritability = 0.80
C. Narrow sense heritability = 0.30
D. Average yield = 430 Units
Explanation:
A. Given that
VA = 30
VD = 50
VI = assumed not available
Therefore
Total genetic variance (VG) = VA + VD
= 30 + 50
= 80
VG = 80
B. Given that
VP = 100
VG = 80
Broad sense heritability, H2 = VG/VP
= 80/100
= 0.80
C. Given that
VA = 30
VP = 100
Narrow sense heritability, h2 = VA/VP
=30/100
= 0.30
D. The difference in selection = 500 - 400
= 100
Recall,
Selection response is heritability multiplied by selection differential.
That is
R = h2S
Selection differential = 100
Heritability h2 = 0.30
Selection response = 0.30 × 100
= 30units
Therefore, expected average yield = 400 + 30
= 430 Units
