Biomolecules Test

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Biomolecules Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following enzymes hydrolyse proteins to $$\alpha$$- amino acids?

A
trypsin

B
pepsin

C
chymotrypsin

D
all of the above

Solution

Enzymes trypsin, pepsin and chymotrypsin hydrolyses proteins to $$\alpha$$- amino acids. Pepsin and trypsin hydrolyzed proteins to produce a complex mixture of peptides called proteoses and peptones. Chymotrypsin cleaves peptide amide bonds and finally breaks proteins into amino acids which is easily absorbed by the body.
Question 2
1 / -0

Which among the following are purine derivatives among the following bases?
1. Adenine 2.Uracil 3.Guanine 4.Thymine

A
1, 4 only

B
2, 3 only

C
1, 3 only

D
1, 3, 4 only

Solution

Adenine and guanine are purine derivatives.
Cytosine and thymine and uracil are pyrimidine derivatives.

So, C is the correct answer.
Question 3
1 / -0

The sequence of bases in the nucleic acid strand is explained by:

A
1$$^{0}$$ structures

B
2$$^{0}$$ structures

C
Both 1$$^{0}$$ and 2$$^{0}$$

D
None

Solution

The sequence of bases in the nucleic acid strand is explained by 1$$^{o}$$ structures. The process of determining the precise order of nucleotides within a $$DNA$$ molecule is called as $$DNA$$ sequencing. It includes any method that is used to determine the order of the four bases $$(AGTC)$$.
Question 4
1 / -0

Glucose reacts with $$CH_{3}OH$$, in presence of dry $$HCl$$ gas, to form:

A
$$\alpha$$-methyl glucoside

B
$$\beta$$-methyl glucoside

C
both A and B

D
none of these

Solution

Glucose reacts with $$CH_{3}OH$$, in presence of dry $$HCl$$ gas, to form a mixture of $$\alpha$$-methyl glucoside and $$\beta$$-methyl glucoside, depending on the approach of the two constituent molecules.
Question 5
1 / -0

Nucleic acids control heredity at:

A
atomic Level

B
ionic level

C
complex level

D
molecular level

Solution

Nucleic acids control heredity at molecular level. The hereditary information is stored as the sequence of bases along the polynucleotide chain. The type, location, and sequencing of the nucleotides control the biological activity of the nucleic acid.
Question 6
1 / -0

Which of the following is not a disaccharide?

A
Cane sugar

B
Raffinose

C
Lactose

D
Maltose

Solution

Cane sugar contains glucose and fructose so it is a disaccharide.

Lactose contains glucose and galactose so it is a disaccharide.

Maltose contains glucose and glucose so it is a disaccharide.

Raffinose contains 1 unit of glucose, 1 unit of fructose and 1 unit of galactose so it is not a disaccharide. It is a trisaccharide.

Option B is correct.
Question 7
1 / -0

Which of the following is least related to the other three?

A
Galactose

B
Glucose

C
Mannose

D
Arabinose

Solution

The compounds galactose, glucose and mannose have 6 carbons while arabinose has 5 carbons in its compound. The image given below is of arabinose.
Question 8
1 / -0

In which of following cases, both the compounds are examples of reducing sugars?

A
Glucose and Sucrose

B
Fructose and Maltose

C
Fructose and Sucrose

D
Sucrose and Lactose

Solution

In reducing sugars, free aldehyde group is present.
Glucose is a reducing sugar and sucrose is a non-reducing sugar.
Fructose, Lactose and Maltose are also reducing sugars.
Question 9
1 / -0

STATEMENT-1 : Sucrose is reducing sugar where as maltose is non reducing sugar.
STATEMENT-2 : In disaccharides, the reducing property depends upon the position of linkages between the two monosaccharide units.

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

C
STATEMENT-1 is True, STATEMENT-2 is False

D
STATEMENT-1 is False, STATEMENT-2 is True
Question 10
1 / -0

Select the correct one from the following order of bonds in $$ A, G, T, C$$ and $$U$$.

A
$$A > G > T > C > U $$

B
$$C = G > T = A = U$$

C
$$G = A = T < C = U $$

D
$$A < G < T < C < U$$

Solution

Order of bonds in $$A, G, T, C$$ and $$U$$ is $$C = G > T = A = U$$.

Within the $$DNA$$ double helix, $$A$$ forms 2 hydrogen bonds with $$T$$ on the opposite strand, and $$G$$ forms 3 hydrogen bonds with $$C$$ on the opposite strand.

Option B is correct.

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Biomolecules Test - 34

Result

Biomolecules Test - 34

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SHARING IS CARING

Question 1

trypsin

pepsin

chymotrypsin

all of the above

Solution

Question 2

1, 4 only

2, 3 only

1, 3 only

1, 3, 4 only

Solution

Question 3

1$$^{0}$$ structures

2$$^{0}$$ structures

Both 1$$^{0}$$ and 2$$^{0}$$

None

Solution

Question 4

$$\alpha$$-methyl glucoside

$$\beta$$-methyl glucoside

both A and B

none of these

Solution

Question 5

atomic Level

ionic level

complex level

molecular level

Solution

Question 6

Cane sugar

Raffinose

Lactose

Maltose

Solution

Question 7

Galactose

Glucose

Mannose

Arabinose

Solution

Question 8

Glucose and Sucrose

Fructose and Maltose

Fructose and Sucrose

Sucrose and Lactose

Solution

Question 9

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

Question 10

$$A > G > T > C > U $$

$$C = G > T = A = U$$

$$G = A = T < C = U $$

$$A < G < T < C < U$$

Solution

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