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Maksim231197 [3]
3 years ago
15

Complete the sentences to correctly describe steps in fatty acid synthesis. move the terms to the appropriate blanks. some terms

will not be used, and some will be used more than once. scroll down to see all the terms. note that a single red x will appear if you answer any of the blanks incorrectly. feedback will be placed next to a blank that is incorrectly filled. fatty acid synthesis begins with a preparatory step in which is transferred from to the . however, it cannot pass through the membrane, so it is transported as , which is cleaved to and . in the cytosol, acetyl coa is converted to , a three-carbon compound. fatty acid synthesis begins with the transfer of the acetyl group from acetyl coa to fatty acid synthase. -carbon groups, supplied by , are added to the growing acyl chain in a series of steps involving condensation, , and dehydration reactions. elongation of the fatty acid chain stops at carbon atoms, after cycles, as the free fatty acid is released.
Biology
1 answer:
saw5 [17]3 years ago
8 0

The synthesis of fatty acids starts with a preparatory step in which acetyl-CoA is mediated from mitochondria to the cytosol. However, it cannot pass through the membrane, so it is transported as citrate, which is cleaved to acetyl CoA and oxaloacetate.

In the cytosol, acetyl CoA is transformed to malonyl CoA, that is, a three carbon compound. Fatty acid synthesis starts with the conduction of acetyl group from acetyl CoA to fatty acid synthase.

Two carbon groups, supplied to malonyl CoA, are supplemented to the developing acyl chain in a series of steps involving condensation, reduction, and dehydration reactions. Elongation of the fatty acid chain ceases at 16 carbon atoms, after seven cycles, as the free free fatty acid is discharged.

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A geneticist crossed fruit flies to determine the phenotypic ratio. The geneticist crossed a fly with blistery wings and spinele
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Complete question:

A geneticist crossed fruit flies to determine the phenotypic ratio. The geneticist crossed a fly with blistery wings and spineless bristles (bbss) with a heterozygous fly that had normal wings and normal bristles (BbSs). Which proportion of offspring that are dominant for both traits in would you not expect based on Mendel's law of independent assortment? 1/2 , 4/16, 25% , or 1/4

Answer:

1/2 is the proportion of the offspring that is NOT expected among individuals that are dominant for both traits.

4/16 = 1/4 = 25% of the progeny and the correct expected proportion of individuals that are dominant for both traits.

Explanation:

<u>Available data</u>:

  • Cross:  a fly with blistery wings and spineless bristles with a heterozygous fly that had normal wings and normal bristles
  • Recessive trait: blistery wings and spineless bristles
  • Dominant trait: normal wings and normal bristles

Let us say that:

  • B is the dominant allele for normal wings
  • b is the recessive allele for blistery wings
  • S is the dominant allele for normal bristles
  • s is the recessive allele for spineless bristles

Parentals)        bbss       x        BbSs

Gametes)  bs, bs, bs, bs     BS, Bs, bS, bs

Punnett square)    BS        Bs         bS        bs

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

F1)  4/16 = 1/4 = 25%  of the progeny is expected to be BbSs, dyhibrid individuals, expressing normal wings and normal bristles

     4/16 = 1/4 = 25% of the progeny is expected to be Bbss, expressing normal wings and spineless bristles

     4/16 = 1/4 = 25% of the progeny is expected to be bbSs, expressing  blistery wings and normal bristles

     4/16 = 1/4 = 25% of the progeny is expected to be bbss, expressing  blistery wings and spineless bristles    

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