150 practice questions covering Biomolecules, organised into 0 topics.
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Sample questions
Q1
medium
Which pair of nitrogen bases are both purines?
Aadenine and guanine
Bcytosine and thymine
Cadenine and thymine
Duracil and cytosine
Show answer and explanation
Correct answer: A - adenine and guanine
Purines are the double ringed bases, and adenine and guanine are the two of them found in nucleic acids. Cytosine, thymine and uracil are all single ringed pyrimidines, so cytosine with thymine and uracil with cytosine are pyrimidine pairs. Adenine with thymine pairs one purine with one pyrimidine, which is a base pair but not a pair of purines.
Q2
medium
Successive nucleotide units in a nucleic acid chain are joined through
Aa peptide linkage between the two phosphate groups
Ba glycosidic linkage between C-1 of one sugar and C-4 of the next
Ca phosphodiester linkage between C-5 of one sugar and C-3 of the next
Da hydrogen bond between the bases of the two units
Show answer and explanation
Correct answer: C - a phosphodiester linkage between C-5 of one sugar and C-3 of the next
One phosphate group forms two ester bonds, one to C-5 of a sugar unit and one to C-3 of the neighbouring sugar, so the backbone of the chain is a series of phosphodiester linkages. A glycosidic linkage from C-1 to C-4 joins sugar units in carbohydrates, hydrogen bonds hold the two strands together rather than building one strand, and a peptide linkage needs an amino group and a carboxyl group.
Q3
hard
Consider these statements about sugars. (i) Ribose is a five carbon monosaccharide. (ii) Every disaccharide releases two identical monosaccharides on hydrolysis. (iii) Inulin is a polysaccharide built from fructose units. Which of them are correct?
A(i) and (ii) only
B(i), (ii) and (iii)
C(i) and (iii) only
D(ii) and (iii) only
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Correct answer: C - (i) and (iii) only
Ribose is indeed a pentose, and inulin is a polymer of fructose, so (i) and (iii) stand. Statement (ii) is wrong: sucrose gives glucose and fructose and lactose gives glucose and galactose, so the two units released need not be the same. Every option that keeps (ii) therefore fails.
Q4
medium
A young child has soft, bent leg bones and spends almost no time in sunlight. The vitamin most likely to be lacking is
Avitamin E
Bvitamin A
Cvitamin $\text{B}_6$
Dvitamin D
Show answer and explanation
Correct answer: D - vitamin D
Vitamin D is needed for bone to harden properly and it is formed in the skin on exposure to sunlight, so a child kept away from the sun develops rickets with soft, deformed bones. Lack of vitamin A causes night blindness, lack of vitamin E makes red blood cells fragile, and lack of vitamin $\text{B}_6$ can bring on convulsions.
Q5
medium
Two monosaccharide units in a disaccharide are held together by
Aa glycosidic bond
Ba peptide bond
Ca phosphodiester bond
Da hydrogen bond
Show answer and explanation
Correct answer: A - a glycosidic bond
Sugar units are joined through their hydroxyl groups by a glycosidic bond, formed with the loss of a water molecule. A peptide bond joins amino acids in a protein and a phosphodiester bond joins nucleotides in a nucleic acid. Hydrogen bonds are weak attractions that hold shapes in place; they are not what links the two sugar units.
Q6
hard
Assertion (A): Every monosaccharide, whether it is an aldose or a ketose, behaves as a reducing sugar. Reason (R): A ketose has no free aldehyde group and therefore cannot reduce Tollens' reagent.
ABoth A and R are true, and R is the correct explanation of A
BBoth A and R are true, but R is not the correct explanation of A
CA is true, but R is false
DA is false, but R is true
Show answer and explanation
Correct answer: C - A is true, but R is false
Every monosaccharide, including the ketose fructose, reduces Tollens' and Fehling's reagents, so (A) is true. (R) is false, because the reagents are alkaline and in alkali a ketose rearranges through an enediol into an aldose, giving the free aldehyde group that is then oxidised.
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