Answer to Question #193611 in Biochemistry for kelvin campbell

Question #193611

1.A nucleic acid has 13% of adenine. Determine the composition of guanine, cytosine

and thymine.

2. What are the forces that stabilize the double stranded DNA?

3. Name the type of bond that links

a. A nucleotide to another nucleotide in a DNA molecule

b. A nitrogenous base to a ribose sugar in an RNA molecule

4. Give the structural differences between RNA and DNA

5. Under aerobic catabolism of glucose, in which compartment of the eukaryotic cell does

the following reactions occur?

a. Conversion of pyruvate to acetyl CoA.

b. Conversion of succinyl-CoA to succinate

c. Conversion of NADH to ATP

d. Conversion of phosphoglycerate to phosphoenolpyruvate

6. Describe various mechanisms for regulation of blood glucose.

7. Give an account of β – oxidation of saturated even carbon fatty acid (Palmitic acid)

along with the total ATPs obtained. 

Expert's answer

(1) According to Chargaff rule, Adenine (A) is paired to Thymine (T) and Cytosine (C ) is paired to Guanine (G).[A]+[G]+[C]+[T]=100% { A=T : G=C }

Here in the question we have [A]=13% therefore % of [T] is also 13%.

Therefore [G]+[C]=100 - 26 = 74 %

Then [G]=37% and [C]=37%

(2) Hydrogen bonds

(3) (a) phosphodiester bonds

b) Glycosidic Bond

(4) The three main structural differences between RNA and DNA are as follows: RNA is single-stranded while DNA is double-stranded. RNA contains uracil while DNA contains thymine. RNA has the sugar ribose while DNA has the sugar deoxyribose.

(5) All the above reactions takes place in the matrix of mitochondria of the cell.

(6) Blood glucose can be regulated by glycolysis, gluconeogenesis, glycogenolysis.

Blood glucose converted into glucagon and stores in liver on action of insulin it converts into glucose.

It maintained by two hormones insulin and glucagon.

(7) Once inside the mitochondria, each cycle of β-oxidation, liberating a two carbon unit (acetyl-CoA), occurs in a sequence of four reactions:

1. Dehydrogenation by FAD: The first step is the oxidation of the fatty acid by Acyl-CoA-Dehydrogenase. The enzyme catalyzes the formation of a double bond between the C-2 and C-3.

2. Hydration: The next step is the hydration of the bond between C-2 and C-3. The reaction is stereospecific, forming only the L isomer.

3. Oxidation by NAD+: The third step is the oxidation of L-β-hydroxyacyl CoA by NAD+. This converts the hydroxyl group into a keto group.

4. Thiolysis: The final step is the cleavage of β-ketoacyl CoA by the thiol group of another molecule of Coenzyme A. The thiol is inserted between C-2 and C-3.

This process continues until the entire chain is cleaved into acetyl CoA units. The final cycle produces two separate acetyl CoAs, instead of one acyl CoA and one acetyl CoA. For every cycle, the Acyl CoA unit is shortened by two carbon atoms. Concomitantly, one molecule of FADH2, NADH and acetyl CoA are formed.

From the Beta oxidation process: The P/O ratios, which give the number of ATP produced per molecule, are 1.5 for FADH2, 2.5 for NADH, and 10 for acetyl-CoA. Therefore, the total ATP yield of oxidation of palmitic acid is 106 ATP.

The total ATP produced can be calculated as follows:

16/2=8, 8×10=80

16/2-1=7, 7×1.5=10.5


The total amount of ATP should be 108 but, 2 ATP molecules are used for the initial activation of every fatty acid that is going to be oxidized in the mitochondria.

Therefore the actual total ATP produced is,

108-2 = 106 ATP

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