Biomolecules Test

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Biomolecules Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

Non-reducing sugar is:

A
maltose.

B
sucrose.

C
lactose.

D
None of these.

Solution

Sucrose is a non-reducing sugar due to the absence of a free aldehyde/ketone group whereas maltose and lactose are both reducing sugars
Question 2
1 / -0

Which set of terms correctly identifies the carbohydrate shown?
(1) Pentose (2) Pentulose (3) Hexulose (4) Hexose (5) Aldose (6) Ketose (7) Pyranose (8) Furanose

A
2,6,8

B
2,6,7

C
1,5,8

D
A set of terms other than these

Solution

The following Carbohydrate is Pentulose (5 membered ring with ketone as a functional group) , Ketose (Contains ketone as the functional group) and Furanose (5 Membered RIng)
So Option A is the answer
Question 3
1 / -0

which of the following pair is $$C_2$$ epimer:

A
D-Glucose , D- Maltose

B
D- Glucose , D- Mannose

C
D- Allose , D- Ribose

D
D-Glucose , D- Arabinose

Solution

D-Mannose suffers from D-glucose only at $$2$$-carbon atom .
$$\Rightarrow$$D-Mannose and D-glucose are $$C_{2}$$-epimers.
Question 4
1 / -0

$$NH_2 - \underset{CH_3}{\underset{|}{C}H} - \overset{O}{\overset{||}{C}} - NH - CH_2 - CO_2H$$
Identify the amino acid obtained by hydrolysis of the above compound.

A
Glycine

B
Alanine

C
Both a and b

D
None of these

Solution
Question 5
1 / -0

D-glucose & D-fructose can be differentiated by :

A
Fehling solution

B
Tollens reagent

C
Benedict test

D
$$Br_2/H_2O$$

Solution

$$Br_2/H_2O$$ being a mild oxidising agent can only oxidise aldose to carboxylic acids but cannot oxidise ketose.

As glucose contains aldehyde group it gets oxidised and converted into a carboxylic acid and is differentiate from fructose which has ketone group.

Therefore, the correct option is D.
Question 6
1 / -0

One cyclic acetal form of D-galactose is shown.
Which atom is the anomeric carbon?

A
Atom A

B
Atom B

C
Atom C

D
Atom D

E
Atom E

Solution

Atom $$F$$ is anomeric in cyclic acetal form of $$D$$ galactose as shown in given figure of question.Anomeric carbon is stereo center anomeric carbon is either $$\alpha,\beta$$ depending on position of $$-OH$$ group.If $$-OH$$ group down in cyclic form then $$\alpha$$ else it is $$\beta$$.
Question 7
1 / -0

Which of the following isomeric sugars is most stable?

A

B

C

D

Solution

We will draw its stable form i.e. chair form. If all the bulky group possess equatorial position then the compound is in stable form.
Some $$-OH$$ group are above the plane and some are below the plane. Thus we will draw accordingly.
Question 8
1 / -0

Which of the following compounds will not undergo decomposition on passing electricity through aqueous solution?

A
Sugar

B
Sodium chloride

C
Sodium bromide

D
Sodium acetate

Solution

Sodium chloride, sodium bromide, and sodium acetate are all ionic compounds and hence, they dissociate into their respective ions in an aqueous solution. On the contrary, sugar is a covalent compound and hence does not dissociate into ions
Question 9
1 / -0

Which glycosidic linkage is present in Maltose?

A
$$\alpha -D-\left( + \right) -glucose\left( { C }_{ 1 } \right) -O-\left( { C }_{ 4 } \right) -\alpha-D-\left( + \right) -glucose$$

B
$$\alpha -D-\left( + \right) -glucose\left( { C }_{ 1 } \right) -O-\left( { C }_{ 2 } \right) -\beta -D-\left( - \right) -glucose$$

C
$$\beta -D-\left( + \right) -glucose\left( { C }_{ 1 } \right) -O-\left( { C }_{ 4 } \right)-\beta -D-\left( + \right) -glucose$$

D
$$\beta -D-\left( + \right) -glucose\left( { C }_{ 1 } \right) -O-\left( { C }_{ 4 } \right) -\alpha-D-\left( - \right) -glucose$$

Solution
Question 10
1 / -0

The two functional groups present in a typical carbohydrate are :

A
$$-CHO$$ and $$-COOH$$

B
$$-CO$$ and $$-OH$$

C
$$-OH$$ and $$-CHO$$

D
$$-OH$$ and $$-COOH$$

Solution

Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones OR compounds which give aldehydic as well as ketonic groups on hydrolysis. Therefore the major functional groups are (carbonyl) Aldehyde, Ketone, and hydroxy groups.