The characteristic of the weak non covalent interactions that enable the assembly, stability and functions of bio molecules is that NON COVALENT INTERACTIONS HAVE LARGE AGGREGATE STRENGTH WHEN MULTIPLE INTERACTION ARE PRESENT.
Although they are considered as weak forces, but when present in large number, non covalent interaction has the capacity to stabilize large polymers using their aggregate strength.
There 2 ways to solve this
1- Directly :
80% of water in the fresh berries è 20% of water in the dried berries
36 Ibs of the fresh berries è X Ibs of the dried berries
X = 
<span> = 9 Ibs</span>
<span>2- </span>If you do not understand the first method, the second one is more explicit:
First, we gonna calculate the weight of the berries without water (dehydrated):
<span>36 * 80% = 28.8 Ibs (of water in the berries)</span>
Then:
<span>36 – 28.8 = 7.2 Ibs (of completely dehydrated berries)</span>
Next, we gonna add 20% of water in the dehydrated berries:
X (total weight of dried berries) è 100%
7.2 Ibs of dehydrated berries è 80%
X = <span> </span>
= 9 Ibs
The correct answer is <span>oil spills.
Bioremeditation is best described as </span><span>a </span>process<span> used to </span>react with different sources, such as contaminated<span> media, </span>inclusive of<span> water, soil and subsurface </span>cloth<span>, through the </span>changing<span> environmental </span>conditions<span> to stimulate </span>growth<span> of microorganisms and degrade the </span>targeted pollutants, in lots of cases<span>, bioremediation is </span>much less highly-priced<span> and </span>greater<span> sustainable than </span>different<span> remediation </span>alternatives<span>. O</span>rganic remedy<span> is a </span>comparable method <span>used to </span>treat<span> wastes </span>consisting of<span> wastewater, </span>industrial<span> waste and </span>solid<span> waste.</span>
Answer:
The correct answers are 2.23 * 10^8 CFU/ml and 2 colonies.
Explanation:
Based on the given information, 0.1 ml is the amount of bacterial culture plated, 10^-5 is the dilution factor and the number of bacterial colonies produced is 223.
A) 223 is the number of colonies produced when 0.1 ml of the culture is plated. Therefore, the number of colonies produced when 1 milliliter of bacterial culture plated us (223/0.1)*1 = 2230
The calculation of the CFU/ml is done by using the formula,
CFU/ml = Number of colonies per ml plated / dilution factor
Thus, 2230/10^-5
= 2230 * 10^5 or 2.23 * 10^8 CFU/ml
B) The number of colonies, which would grow on a plate, which is inoculated with 0.1 ml volume of 10^-7 dilution from the similar bacterial stock will be calculated as,
CFU/ml = Number of colonies per ml plated/ dilution * volume plated.
2.23 * 10^8 CFU/ml = Number of colonies per ml plated / 10^-7 * 0.1
Number of colonies per ml plated = 2.23 * 10^8 * 0.1 / 10^7 = 2.23 or 2 colonies.
Gene transfer to nontarget species
