Answer:
Inputs for the Electron Transport Chain are NADH, H+, ADP, FADH2, O2. Outputs: NAD+, ATP, FAD, H2O. 1)
Answer:
look at the pictures
Explanation:
1. If we cross the parental generation which is RR x rr, the probability of the F1 generation to get rough coat is 100% and they will be a carrier of the recessive trait smooth coat.
Since the phenotypic ratio is 100% heterozygous Rr, in crossing the F1 to get the F2, we will use the genotype Rr. To get the F2, use the cross Rr x Rr.
The phenotypic ration for F2 is 3:1. There is 75% to get rough coat and 25% smooth. The answer is based on the result on the Punnett square. On the other hand, the genotypic ratio is 1:2:1. There is 25% probability to get RR genotype, 50% Rr, and 25% rr.
2. Since the two parental mice got 6 albino offspring and 5 brown mice offspring, it is approximately 50%. it takes a parental mice who is Brown that is carrying an albino trait crossed with an albino to get offspring with almost the same number. Therefore, the genotype of the brown mice is Aa.
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Answer:
The reduced form of cytochrome c more likely to give up its electron to oxidized cytochrome a having a higher reduction potential.
Explanation:
Electrons from NADH and FADH2 flow spontaneously from one electron carrier of the electron transport chain to the other. This occurs since the proteins of the ETC are present in the order of increasing reduction potential. The reduced cytochrome b has lower reduction potential than cytochrome c1 which in turn has a lower reduction potential than the cytochrome c.
Cytochrome c is a soluble protein and its single heme accepts an electron from cytochrome b of the Complex III. Now, cytochrome c moves to complex IV which has higher reduction potential and donates the electron to cytochrome a which in turn passes the electrons to O2 via cytochrome a3.