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Triss [41]
2 years ago
15

What was the biochemical reason for the "Ox Phos War," i.e., why was Peter Mitchell's hypothesis considered heretical

Biology
1 answer:
Mnenie [13.5K]2 years ago
6 0

Answer:

The options

A ) It was thought that all ATP in the cell was synthesized from the glycolytic pathway in the cytosol and that mitochondria are dispensable.

B) It was thought that oxidative phosphorylation required a reactant with a high energy bond (~X) and it was just a matter of who could find it first.

C) The reaction A~X + Pi + ADP. A + X + ATP was thought to be the preferred pathway using both a membrane potential and Na+ gradient.

D) Since Peter Mitchell was a wealthy scholar who lived in a mansion with a built-in laboratory, it was assumed he was eccentric and wrong.

The CORRECT ANSWER IS B

B) It was thought that oxidative phosphorylation required a reactant with a high energy bond (~X) and it was just a matter of who could find it first.

Explanation:

Majority of the biochemists at that period were experimenting to evaluate for a high energy phosphorylated intermediate also known as ~X-P and they believed it is the answer to the question or experiment.

Peter Mitchell's hypothesis was considered heretical because it was independent on ~X-P. His model was considered unusual and a new model plus he had his own private laboratory... a factor wasn't common too at that period.

It was called "Ox Phos War due to the disagreement "war" as biochemists had to chose sides.

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Hello there,

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A boy threw a small bundle toward his girlfriend on a balcony 10. meters above him. The bundle stopped rising in 1.5 seconds. Ho
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Hi!


We know that the boy must have thrown the bundle with some initial velocity 'v', and that the bundle stops after 1.5 seconds at a height 's' with a final velocity 'u' -which would be 0.

Our known values thus are:

Time (t) = 1.5 s

Final velocity (u) = 0 m/s

Gravitational acceleration (g) = -9.8 m/s^(2)

<em>Note: that the acceleration is negative as it is opposing the direction of motion (acting  down on the object, causing deceleration)</em>

We will be using <em>equations of motion for an accelerating object</em> to solve first for initial velocity, in order to find the distance.

<h3>Initial Velocity using     <u>v = u + at</u></h3>

Rearranging the equation, and using the value 'g' as our acceleration, we get,<em> u = v - at</em>

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<h3>Distance Travelled using   <u>v^2 = u^2 + 2as</u></h3>

Rearranging the equation, we get, <em>s= v^(2) - u^(2) / 2a</em>

s = 0^(2) - 14^(2) / 2(-9.8)                          

<em>Note: the negative sign is not a part of the initial velocity, but a part of the equation </em>

s = - 216.09 / -19.6 = 11.025 m


The distance traveled by the bundle is hence, ~ 11 meters, which is high enough for the girl to catch.


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