Biology question- hydrolysis

13

Hi, please could I have help on this question? I know that hydrolysis is required so statement 1 is definitely correct as heat and enzymes are present. I’m not really sure about the other two statements though?
Here is the question: https://ibb.co/H24Lj74
Thanks!

Reply 1

Nitrotoluene

12

Original post

by anonymous56754

Hi, please could I have help on this question? I know that hydrolysis is required so statement 1 is definitely correct as heat and enzymes are present. I’m not really sure about the other two statements though?
Here is the question: https://ibb.co/H24Lj74
Thanks!

1) The procedure works: If reducing sugars are present, the test will show a colour change, indicating their presence.
2) The procedure does not work: The polymer will not break down into its monomer units when dissolved in water and neutralized. Benedict's reagent will only react with the polymer if it possesses reducing ends. This does not apply to your 2nd question.
3)The procedure works: If reducing sugars are present, a colour change will occur when the solution is neutralized and then heated with Benedict's reagent.
Therefore, the correct answer is C. 1 and 3 only.
Ciao,
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Sandro
My signature: "Regardless of where you may be, expressing gratitude is a universally cherished gesture."

Reply 2

anonymous56754

OP

13

Original post

by Nitrotoluene

1) The procedure works: If reducing sugars are present, the test will show a colour change, indicating their presence.
2) The procedure does not work: The polymer will not break down into its monomer units when dissolved in water and neutralized. Benedict's reagent will only react with the polymer if it possesses reducing ends. This does not apply to your 2nd question.
3)The procedure works: If reducing sugars are present, a colour change will occur when the solution is neutralized and then heated with Benedict's reagent.
Therefore, the correct answer is C. 1 and 3 only.
Ciao,
The flag of Italy.png

Sandro
My signature: "Regardless of where you may be, expressing gratitude is a universally cherished gesture."

Thank you for your reply, but I’m still quite confused about statements 2 and 3? For statement 2, is the heat not sufficient to break the glycosidic bond and produce a reducing sugar? Also does the HCl acid cause hydrolysis (which is why it doesn’t work with 2 because there is no HCl?)

Reply 3

username7368346

16

1.

Heating e.g. starch certainly weakens internal glycosidic bonds (gelatinisation at 60–80 degrees) and the polymer will sooner or later start to break apart. Like you I suppose that a polymer is not heated long enough in a simple Benedict’s test — the quoted temperature in my notes is 90 degrees — to break down.

2.

Yes, acid hydrolysis — the equation for the non-reducing disaccharide sucrose is:

If the screenshot doesn’t work, C12H22O11 + H2O => C6H12O6 (glucose) and C6H12O6 (fructose), which of course isomers, with glucose and fructose both being Benedict’s reacting reducing sugars.

The HCl acts as a catalyst — the hydroxonium ions it forms in aqueous solution act as proton donors to sucrose molecules, producing a more reactive intermediate molecule that hydrolyses far more readily. The various rearrangements regenerate a hydroxonium ion as well as the two monosaccharides. I have notes on the acid hydrolysis of an ester but as I didn’t do IB I don’t think I have any reaction diagrams for acid hydrolysis of a polysaccharide.

Reply 4

TypicalNerd

18

Original post

by anonymous56754

Thank you for your reply, but I’m still quite confused about statements 2 and 3? For statement 2, is the heat not sufficient to break the glycosidic bond and produce a reducing sugar? Also does the HCl acid cause hydrolysis (which is why it doesn’t work with 2 because there is no HCl?)

Statement 2 says something about neutralisation- what exactly is being neutralised?

Also, simply heating a sugar in solution quite probably won’t break it down. It won’t get hot enough (most polysaccharide sugars decompose well above the boiling point of water and usually form chemically very different products to the monosaccharides - these probably wouldn’t respond to benedict’s test if they were to be produced).

Reply 5

Nitrotoluene

12

Original post

by TypicalNerd

Statement 2 says something about neutralisation- what exactly is being neutralised?
Also, simply heating a sugar in solution quite probably won’t break it down. It won’t get hot enough (most polysaccharide sugars decompose well above the boiling point of water and usually form chemically very different products to the monosaccharides - these probably wouldn’t respond to benedict’s test if they were to be produced).

Hello Typical Nerd!
IMHO, neutralisation is an important step, if required, because the Benedict reagent reacts better in a neutral or slightly alkaline environment than in an acidic or alkaline environment.
Supplement: I checked my textbooks and found that some enzymes such as invertase or amylase can be acidic. When dissolved in water, these compounds release protons (H^+ ions) and lower the pH of the solution reaching pH < 7.
Ciao,
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Sandro

My signature: "Regardless of where you may be, expressing gratitude is a universally cherished gesture."

Reply 6

TypicalNerd

18

Original post

by Nitrotoluene

Hello Typical Nerd!
IMHO, neutralisation is an important step, if required, because the Benedict reagent reacts better in a neutral or slightly alkaline environment than in an acidic or alkaline environment.
Ciao, The flag of Italy.png

Sandro

My signature: "Regardless of where you may be, expressing gratitude is a universally cherished gesture."

Yes, neutralisation is important - as you have said if you’ve thrown something like concentrated HCl into the solution in advance. In the method described in statement 2, there is no such reagent thrown in advance hence there is no need to neutralise anything. This was my point and I do hope it was generally clear. If not, then thank you for prompting me to clarify.

In fact, in trying to neutralise the solution when it is in fact already neutral, you are actually making the solution less suitable to test with Benedict’s reagent (not that it would matter in this case anyway, since there are no oxidisable aldehydes present).