Answer:
To understand the utility in sequence comparison and in the search for proteins that have a common evolutionary origin, you need to be clear about some concepts about how to evolve proteins. The idea that is accepted is that throughout the evolution some species are giving rise to new ones. Behind this is the genetic variation of organisms, that is, the evolution of genomes and their genes, as well as the proteins encoded by them.
Explanation:
Three ways can be distinguished by which genes evolve, and by proteins: mutation, duplication and shuffling of domains. When differences between homologous protein sequences are observed, these differences change to do with the way of life of the organism, an example of this, bacteria that live in hot springs at very high temperatures have proteins with a very high denaturation temperature, and these proteins are usually richer in cysteines. On the other hand, the fact that in positions of the sequences they remain unchanged (preserved positions), means that these have a special importance for the maintenance of the structure or function of the protein and its modification has not been tolerated throughout of evolution
Answer:
PH= 6.767 (answer is the A option)
Explanation:
first we need to correct the value in Kw at this temperature is 2.92*10^-14
so, in this case we have that:
Kw=2.92*10^-14 M²
[ H3O^+] [ H3O^+]
![[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%5BOH%5E%7B-%7D%20%20%5D%20%3D%20Kw%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D%20%20%20%5C%5C%5C%5C)
at 40ºC
![[H_{3}O^{+} ] = [OH^{-} ]](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%5BOH%5E%7B-%7D%20%20%5D)
![[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%5E%7B2%7D%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D)
![[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%282.92%2A10%5E%7B-14%7D%29%5E%7B1%2F2%7D%20%3D%201.71%2A10%5E%7B-7%7D%20M)
![PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767](https://tex.z-dn.net/?f=PH%3D%20-log10%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20-log10%281.71%2A10%5E%7B-7%7D%20%29%20%3D%206.767)
Answer:
D. 0.160
Explanation:
The solution A is obtained adding 2.0mL of a solution of bromocresol green, 5.0mL of 1.60M HAc and 2.0mL of a solution of KCl. The solution is diluted to 50mL
That means the HAc is diluted from 5.0mL to 50.0mL, that is:
50.0mL / 5.0mL = 10 times.
And the final concentration of HAc must be:
1.60M / 10 times =
0.160M
Right answer is:
<h3>D. 0.160</h3>
Answer:
1.36
Explanation:
HClO2 ⇄ H+ + ClO2-
[HClO2] [H+} [ClO2-]
initial: 0.12 0 0
change: -x +x +x
equil: 0.12-x x x
[H+][ClO2-] / [HClO2] = Ka = 1.2 * 10^-2
x* x / 0.12 - x = 1.2 * 10^-2
x^2 = (2.4 * 10^-3) - (1.2 * 10^-2)x
x^2 + (1.2 * 10^-2)x – (2.4 * 10^-3) = 0
Upon solving for x,
x = 0.04335
[H+] = 0.04335
pH = -log[H+]
pH = -log(0.04335) = - (-1.36) = 1.36
