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zubka84 [21]
3 years ago
14

Free palmitate is activated to it's coA derivative(palmitoyl-CoA) in the cytosol before it can be oxidized in the mitochondria.

If palmitate and radioactively labeled CoA are added to liver slices, palmitoyl-CoA isolated from the cytosolic fraction is radioactive but that isolated from the mitochondrial fraction is not. Explain.
Biology
1 answer:
n200080 [17]3 years ago
7 0

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).

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Answer:

<h2>e) 3/4 </h2>

Explanation:

As given;

Black (B) is dominant over red (b), so B is dominant over b.

Solid (S) is dominant over spotted color(s), so S is dominant over s.

Offspring of BBss and bbss individuals are:

Gametes of parent BBss are Bs, Bs and gamete of bbss are bs, bs

offspring are : all Bbss

Cross between these offspring is; Bbss×Bbss

Gametes of parent Bbss are: Bs and bs  as given that the genes are linked.

Bs, bs × Bb, bs

Offspring are = BBss, Bbss, Bbss, bbss

Black spotted (B_ss) = 3/4

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Hope it helped,


BioTeacher101

3 0
3 years ago
Read 2 more answers
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