Answer:
The cytochrome b6f is a large multi-subunit protein, which accepts electrons from the plastohydroquinone.
One electron moves linearly toward plastocyanin while the other goes through a cyclic process, which effectively pumps more protons into the thylakoid lumen.
Explanation:
The cytochrome b6f is distributed among both grana and stroma thylakoids equally. They are usually large and embedded in the membrane.
Answer:
Option C, he lost both sensory and short-term memory
Explanation:
Sensory memory is basically the information received by the five sense organs and also the instruction given by brain to the respective sense organ.
This sensory memory keeps storing and is collectively referred to as short term memory.
Since Chuck Wildlee was unable to remember how many balls and strikes the batter had. In past he would have perceived this information through eyes which should have been part of sensory memory and later the part of short term memory.
Thus, it can be said that Chuck Wildlee lost both his sensory memory and short term memory
Hence, option C is correct
Answer:
The cytosolic and mitochondrial pools of CoA are kept separate, and no radioactive CoA from the cytosolic pool enters the mitochondrion.
Explanation:
- Fatty acyl group condensed with CoA in the cytosol are first transferred to carnitine and in this process, CoA is released.
- After this, it is transported into the mitochondrion, where it is again condensed with CoA.
- In this way, the cytosolic and mitochondrial pools of CoA are kept separate, and due to this reason, no radioactive CoA from the cytosolic pool enters the mitochondrion.
- Therefore, according to the given question, the C14 CoA that is added into the liver homogenate along with palmitate shows cytosolic radioactive fraction but not mitochondrial as in the mitochondria a different CoA joins palmitate and not the one containing C14.
Answer:
D
Explanation:
D is really decomposing waste of anytime so I think that’s the answer.
Answer:
Explanation:
From the question, the total population = 171 i.e (80+71+21)
Genetic frequency therefore; BB = 71/171 = 0.42
Bb = 80/171 = 0.47
bb = 20/171 = 0.12
Allele frequency therefore;
Allele B = 0.42+(0.50×0.47) = 0.66
Allele b = (0.50×0.47)+0.12 = 0.36
Note all values were rounded up to two significant figures
