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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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A geneticist crossed pure breeding black mice with pure breeding brown mice. All the 992 mice in F1 generation had black coats. When these mice were crossed, the yielded 961 black coated mice and 317 brown coated mice. How could you account for the ratio of black coated to brown coated mice in the F2 generation? Select all that apply

  1. The F1 male produced two kinds of sperm, half carrying an allele for black coat color, and half carrying an allele for brown coat color.
  2. The brown allele segregates together with the black allele and disappears in the F1 generation.  
  3. The F1 female produced two kinds of eggs, half carrying an allele for black coat color, and half carrying an allele for brown coat color.
  4. An approximately three-to-one ratio of black to brown coated mice in the F2 generation is accounted for by the black allele being dominant over the brown allele.

Answer:      

  1. The F1 male produced two kinds of sperm, half carrying an allele for black coat color, and half carrying an allele for brown coat color. APPLIES
  2. The brown allele segregates together with the black allele and disappears in the F1 generation.  DOES NOT APPLY
  3. The F1 female produced two kinds of eggs, half carrying an allele for black coat color, and half carrying an allele for brown coat color. APPLIES
  4. An approximately three-to-one ratio of black to brown coated mice in the F2 generation is accounted for by the black allele being dominant over the brown allele. APPLIES

Explanation:

You will find the complete question, answer, and explanation in the attached files, due to technical problems.  

Download pdf
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