All categories

2 years ago

Identify the true statements regarding alpha-1,6 linkages in glycogen.

Chemistry

1 answer:

frozen [14]2 years ago

3 0

Answer:

a. New alpha- 1,6 linkages can only form if the branch has a free reducing end

b. The number of sites for enzyme action on a glycogen molecule is increased through alpha- 1,6 linkages

c. At least four glucose residues separate alpha-1,6 linkages

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

Explanation:

Glycogen i is the main storage polysaccharide in animals. It a homoplymer of (alpha-1-->4)-linked subunits of glucose molecules, with alpha-1--->6)-linked branches.

The alpha-1,6 branches are formed by the glycogen-branching enzyme which catalyzes the transfer of about 7 glucose residues from the non-reducing end of a glycogen branch having at least 11 residues to the C-6 hydroxyl group of a glucose residue which lies inside the same glycogen chain or another glycogen chain, thereby forming a new branch. This ensures that there are at least four glucose residues separating alpha-1,6 linkages.

The effect of branching is that it makes the glycogen molecule more soluble and also increases the number of non-reducing ends, thereby increasing the number of sites for the action of the enzymes glycogen phosphorylase and glycogen synthase.

You might be interested in

The noble gases are inert. This means they a. exist as gases at room temperature. b. undergo many chemical reactions. c. lose an

eduard

Answer:

Explanation:

The noble gases are said to be inert because they undergo very few chemical reactions. They enjoy their status by having completely filled electron shells and do not have any reason whatsoever to go into chemical reactions. Most elements otherwise go into chemical reactions principally to enjoy the stability of these elements.

Noble gases can be found in the last group of the periodic table.

8 0

2 years ago

a 0.5674 g piece of copper is added to 10.00 ml; of 16 M HNO3, producing 0.8024 g of copper (ii) nitrate. What is the precent yi

uysha [10]

Idk there just making me answer it

5 0

2 years ago

Read 2 more answers

Consider a process in which the entropy of a system increases by 125 J K−1 and the entropy of the surroundings decreases by 125

expeople1 [14]

Answer : The process is not spontaneous.

Explanation :

As, we know that:

Change in entropy = Change in entropy of system + Change in entropy of surrounding

As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.

For the given reaction to be spontaneous, the total change in entropy should be positive.

Given :

Entropy change of system = +125J/K

Entropy change of surroundings = -125J/K

Total change in entropy = Entropy change of system + Entropy change of surroundings

Total change in entropy = 125 J/K + (-125 J/K)

Total change in entropy = 0

The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.

4 0

2 years ago

what is 31/50000 If the toxic quantity is 1.5 g of ethylene glycol per 1000 g of body mass, what percentage of ethylene glycol i

Sveta_85 [38]

Answer:

$\%m=0.15\%$

Explanation:

Hello,

In this case, we are asked to compute the by mass percent representing the toxicity of ethylene glycol in the body mass. In such a way, since the by mass percent is computed with the shown below formula:

$\%m=\frac{m_{ethylene \ glycol}}{m_{ethylene \ glycol}+m_{body\ mass}}*100\%$

We can use the given masses to obtain:

$\%m=\frac{1.5g}{1.5g+1000g}*100\%\\ \\\%m=0.15\%$

Best regards.

7 0

2 years ago

A species has the following MO configuration: (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(σ2p)2(π2p)2. This substance is:_______.

Goshia [24]

Answer : The correct option is, (a) paramagnetic with two unpaired electrons.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 14 electrons present in the given configuration.

The molecular orbital configuration of molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^1=(\pi_{2p_y})^1],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The number of unpaired electron in the given configuration is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.

Hence, the correct option is, (a) paramagnetic with two unpaired electrons.

3 0

2 years ago