4. If cells synthesizing glucose from lactate are exposed to CO 2 labeled with 1

Question

4. If cells synthesizing glucose from lactate are exposed to CO2 labeled with 14C, what will be the distribution of label in the newly synthesized glucose?

C-1 and C-6

C-2 and C-5

C-3 and C-4

C-1 and C-4

None of the answers is correct.

5. In prokaryotes the site of ATP-synthesizing machinery is the:

mitochondrial matrix.

outer cell wall.

cytoplasmic membrane.

nucleolus.

None of the answers is correct.

–33.5 kJ/mol

+33.5 kJ/mol

–14.2 kJ/mol

+14.2 kJ/mol

–57.7 kJ/mol

C-1 and C-6

C-2 and C-5

C-3 and C-4

C-1 and C-4

None of the answers is correct.

Explanation / Answer

Ans. 4. E. None of the answers is correct.

Lactate is converted back into pyruvate through Cori cycle in hepatocytes as follows-

            2 Lactate -------> 2 Pyruvate ----+6 ATP---> Glucose

No CO2 is incorporated into glucose in Coir cycle. Therefore, there 14C is present in glucose.

Ans. 5. C. Cytoplasmic membrane.

The ATP synthesizing machinery in prokaryotes in embedded in the cytoplasmic membrane (plasma membrane).

Mitochondria and nucleolus are absent in bacteria.

Outer cell wall does not ATP synthesizing machinery.

Ans. 6. Acetate + ATP + CoA ----> Acetyl-CoA + AMP + 2Pi      - reaction 1

Given,

ATP + H2O ----> AMP + PPi ,                       dGo’ = -45.6 kJ mol-1    - reaction 2

Acetyl-CoA + H2O ----> Acetate + CoA        , dGo’ = -31.4 kJ mol-1 - reaction 3

PPi + H2O ----> 2 Pi                                       , dGo’ = -19.3 kJ mol-1 - reaction 4

Now,

            Acetate + CoA ----> Acetyl-CoA + H2O        reverse of reaction 3

+         ATP + H2O     ----> AMP + PPi                     reaction 2

+         PPi + H2O       ----> 2 Pi                                 reaction 4

------------------------------------------------------

=          Acetate + CoA + ATP ­----> Acetyl-CoA + AMP + 2 Pi

Using Hess’ Law of constant heat summation,

dGo’ (reaction 1) = dGo’ (reverse of reaction 3) + dGo’ (reaction 2) + dGo’ (reaction 4)     

or, dGo’ (reaction 1) = +31.4 kJ mol-1   + (- 45.6 kJ mol-1   ) + (-19.3 kJ mol-1 )

or, dGo’ (reaction 1) = -33.5 kJ mol-1

Hence, dGo’ = -33.5 kJ mol-1

Correct option = B

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