g Identify the true statements regarding α-1,6 linkages in glycogen. At least four glucose residues separate α‑1,6 linkages. The number of sites for enzyme action on a glycogen molecule is increased through α‑1,6 linkages. New α‑1,6 linkages can only form if the branch has a free reducing end. The reaction that forms α-1,6 linkages is catalyzed by a branching enzyme. Exactly four residues extend from these linkages.

At least four glucose residues separate α‑1,6 linkages.

The number of sites for enzyme action on a glycogen molecule is increased through α‑1,6 linkages.

The reaction that forms α-1,6 linkages is catalyzed by a branching enzyme.

Explanation:

Glycogen is a polymer of glucose and is the primary carbohydrate storage form in animals. The polymer is composed of glucose units linked in alpha(1-4) straight chains and alpha(1-6) branches which occur on average every 8-12 straight chain glucose residues. It has a reducing and non-reducing end. The end of the molecule containing a free carbon number one on glucose is called a reducing end. The other ends are all called non-reducing ends.

During the breakdown of glycogen, glucose units are removed one at a time from the non-reducing end until a point about four glucose residues away from a branch which will require a debranching enzyme to act for further breakdown to occur. Since many such branches occur in a glycogen molecule, it makes it possible for breakdown of glycogen to occur at many points speedily.

Glycogen branching enzyme is required to make alpha (1-6) glycosidic bonds. It transfers 6 to 7 glucose units from the non-reducing end of a straight chain glycogen molecule to an interior position of the same or another glycogen molecule forming alpha (1-6) bonds.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml