Answer:
14 CO₂ will be released in the second turn of the cycle
Explanation:
<u>Complete question goes like this</u>, "<em>The CO2 produced in one round of the citric acid cycle does not originate in the acetyl carbons that entered that round. If acetyl-CoA is labeled with 14C at the carbonyl carbon, how many rounds of the cycle are required before 14CO2 is released?</em>"
<u>The answer to this is</u>;
- The labeled Acetyl of Acetyl-CoA becomes the terminal carbon (C4) of succinyl-CoA (which becomes succinate that is a symmetrical four carbon diprotic dicarboxylic acid from alpha-ketoglutarate).
- Succinate converts into fumarate. Fumarate converts into malate, and malate converts into oxaloacetate. Because succinate is symmetrical, the oxaloacetate can have the label at C1 or C4.
- When these condense with acetyl-CoA to begin the second round of the cycle, both of these carbons are discharged as CO2 during the isocitrate dehydrogenase and alpha-ketoglutarate dehydrogenase reactions (formation of alpha-ketoglutarate and succinyl-CoA respectively).
Hence, 14 CO₂ will be released in the second turn of the cycle.
First calculate the probability of type AB, which is the remainder after subtracting types A, B and O.
P(AB) = 1-(0.34+0.12+0.5) = 1-0.96 = 0.04
Anti-b will react with types B and AB, so
P(reaction) = P(B)+P(AB) = 0.12 + 0.04 = 0.16
Answer: For this person, the probability of reaction with anti-b is 0.16
The answer to the first question is a.
