Answer:
Lots of bacteria are present as the microflora of a baby's gut which helps it to extract the nutrition from mother's milk and also helps the baby to maintain a low number of pathogenic bacteria in its gut.
Some bacteria present in the baby gut release enzymes like glycoside hydrolase which helps baby to digest carbohydrate present in the mother's milk.
Bacteria like Bifidobacterium, Streptococcus, Staphylococcus are already present in breast milk which helps in making gut microflora of baby and this gut microflora helps in extracting nutrients from mother's milk.
Answer: hydrolysis of intermediate palmitoyl Co A ,with loss of labeled CoA.before reaching the matrix gives the answer
Explanation:
This is because when the labeled Coenzyme A of the Plamitate combines as Palmitoy-CoA with oxaloacetate to form intermediate (palmitoyl-CoA )in Citric Acid cycle:
CoA is hydrolysed with loss of the labelled CoA which returns to the cystosol. Therefore, the labelled CoA does not reach the matrix of the mitochondrial,but returns to the Cystosol.
Consequently, the CoA in the Cystosol will be labelled in palmitoylCoA and the one in the matrix of the liver mitochondrial will be non radioactive(,will not labelled).
Answer:
Semi-conservative replication
Explanation:
After the double-helix discovery of Watson and Crick, there were three possible models about the DNI replication:
- The Conservative model stated that the two strands of DNI together were the template of another new molecule. The final product was the original double-stranded molecule and the new molecule.
- The semi-conservative model stated that the original DNI molecule separated into two strands, and each of them served as a template for the synthesis of a new complementary strand. The replication product would be two double-stranded DNA molecules, each carrying an original strand a new one.
- The Dispersive moles stated that the replication product would be two molecules made by a mixture of segments of the original and the new molecules.
Meselson and Stahl joined to discover which of the models was the correct one. To do it they used E. coli and Nitrogen isotopes.
- First, they extracted DNI from bacteria grown in a medium with N¹⁴ and got its density band by centrifugation.
- Then they grew bacteria in a medium with N¹⁵, extracted their DNI molecules, centrifugated them, and got the density band, which was heavier than the firsts ones.
- The researchers then transferred bacteria grown in medium with N¹⁵ to a medium with N¹⁴, and they allowed only one replication process to occur. DNI was extracted and centrifugated again, and a new band appeared. This band was an intermediate form between bands of DNI-N¹⁵ and DNI-N¹⁴.
This event <em>eliminated the conservative model</em>. If this model were correct, the expected result would be to get two bands: one corresponding to the density DNI-N¹⁵ and the other corresponding to the density DNI-N¹⁴.
- Bacteria grown in a medium with N¹⁵ and then transferred to a medium with N¹⁴ were finally allowed to replicate twice. Their DNI was extracted and centrifugated. The result was two bands: one of them coincided with the intermediate band, and the other one with the DNI-N¹⁴.
<u>This result was conclusive</u> because if the dispersal model were correct, these two bands should not appear, as all the DNI strands would have part of the original molecule.
With this experiment, Meselson and Stahl proved that the correct replication model was the semi-conservative one.
