Answer:
B) FADH2 -- FMN of Complex I -- Fe-S of Complex II -- Q -- Fe-S of Complex III -- Cyt c -- Cyt a of Complex IV -- O2
Explanation:
FADH2 and NADH give their high energy electrons to the terminal electron acceptor molecular oxygen via an electron transport chain. As the electrons move through electron carriers of the electron transport chain, they lose their free energy. Part of the free energy of the electrons is used to pump the protons from the matrix into the intermembrane space. Therefore, part of the energy of electrons is temporarily stored in the form of a proton concentration gradient.
NADH gives its electrons to FMN of complex I while FADH2 gives its electrons to the Fe-S center of complex II. Both the complexes are oxidized by coenzyme (Q) which in turn reduces Fe-S centers of complex III. Cyt c of complex IV obtains electrons from complex III and passes them to CuA center, to heme "a" to heme "a3-CuB center" and finally to the molecular oxygen.
So, the compounds arranged with respect to the energy content of electrons in descending order are as follows: FADH2 -- FMN of Complex I -- Fe-S of Complex II -- Q -- Fe-S of Complex III -- Cyt c -- Cyt a of Complex IV -- O2.
<span>The best answer to the question: "The destruction of salt marshes will directly harm each organism except" ... is Algae. Algae blooms are the biggest outcome of salt marsh destructions, which starve the water of oxygen making it uninhabitable, as the algae grows, it covers the surface of the water and harms almost every organism part of that ecosystem. So I would say algae.</span><span />
Surprisingly, nothing.
Without changing the characteristics of the light bulb, it's not actually possible to change the voltage without changing the current as well. When the voltage is increased, the current through the bulb has to increase as well in order to make it glow brighter.
What he said ^^. . . ....... . .
Answer:
Barr bodies are small bodies that are dark staining bodies that show the inactivated X chromosome, these sex chromatin bodies attach to the membrane of the nucleus of interphase cells.
One less of the total number of X-chromosome (n-1) represents the number of Barr bodies.
Then,
The number of Barr bodies with Klinefelter would be:
Klinefelter (47,XXY): (n-1) = (2-1) = 1 Barr body as here two X chromosome present.
Similarly, for
Turner (45,XO): (n-1) = (1-1) = 0 Barr bodies (X chromosome = 1)
Now, as per the question, the bar bodies for karyotypes given
47,XYY: (n-1) = (1-1) = 0 Barr bodies
47,XXX: (n-1) = (3-1) = 2 Barr bodies
48,XXXX: (n-1) = (4-1) = 3 Barr bodies
