Home / iGCSE Biology (0610)-17.1 Chromosomes, genes and proteins – iGCSE Style Questions Paper 4

iGCSE Biology (0610)-17.1 Chromosomes, genes and proteins – iGCSE Style Questions Paper 4

IGCSE Biology (0610) Practice Questions Paper 4 - All Chapters
Question

Fig. 2.1 is a diagram of part of a DNA molecule.

 

(a) Draw a circle on Fig. 2.1 to identify one pair of bases.

(b) The percentage of T bases in the DNA of a species is 29%.

Calculate the percentage of bases that would be base G in the DNA of this species.

Space for working.

(c) (i) State the name given to a length of DNA that codes for a protein.

(ii) Explain how proteins are made by a cell.

(iii) DNA controls cell function by controlling the production of proteins.

State two types of cell membrane proteins.

▶️ Answer/Explanation
Solution

(a)

The student should circle one pair of the inward-pointing horizontal sections opposite each other on the DNA diagram. These represent the complementary base pairs (A-T or C-G).

(b) 21%

Explanation:

According to Chargaff’s rules in DNA:

  • Adenine (A) pairs with Thymine (T), so if T is 29%, then A must also be 29%.
  • This accounts for 58% of the bases (29% A + 29% T).
  • The remaining 42% must be equally divided between Cytosine (C) and Guanine (G) since they pair together.
  • Therefore, G = 42% ÷ 2 = 21%.

(c)(i) gene

A gene is a segment of DNA that contains the instructions for making a specific protein.

(c)(ii)

Explanation of protein synthesis:

  1. Transcription: In the nucleus, the DNA code for a specific gene is copied into messenger RNA (mRNA). This process is called transcription.
  2. mRNA processing: The mRNA molecule is modified and then leaves the nucleus through nuclear pores.
  3. Translation: The mRNA attaches to a ribosome in the cytoplasm. Transfer RNA (tRNA) molecules bring specific amino acids to the ribosome based on the mRNA code.
  4. Peptide bond formation: The ribosome joins the amino acids together with peptide bonds, forming a polypeptide chain.
  5. Protein folding: The polypeptide chain folds into its specific three-dimensional shape to become a functional protein.
  6. Energy requirement: The entire process requires energy in the form of ATP.

(c)(iii) Two types of cell membrane proteins:

  1. Carrier proteins: These help transport molecules across the cell membrane, often against concentration gradients (active transport).
  2. Receptor proteins: These detect and respond to chemical signals from outside the cell, triggering specific cellular responses.
  3. Other possible answers: Channel proteins, enzyme proteins, cell recognition proteins, or structural proteins.
Question

(a) Proteins are biological molecules. Explain how proteins are made.

(b) Fig. 3.1 is a diagram of enzyme 1 catalysing the breakdown of molecule A into two smaller molecules, B and C.

Molecule D, shown in Fig. 3.2, is added to the mixture shown in Fig. 3.1. This causes the rate of production of molecules B and C to decrease.

(i) Using the information in Fig. 3.1 and Fig. 3.2, suggest why the rate of production of molecules B and C decreases after molecule D is added.

(ii) Fig. 3.3 shows enzyme 1 after a change in its environmental conditions.

Suggest two changes that could cause the effect on enzyme 1 shown in Fig. 3.3

▶️ Answer/Explanation
Solution

(a) Protein Synthesis:

  1. Transcription: mRNA is synthesized in the nucleus as a complementary copy of a DNA gene.
  2. mRNA Processing: The mRNA strand exits the nucleus and attaches to a ribosome in the cytoplasm.
  3. Translation: The ribosome reads the mRNA sequence in codons (3-base groups).
  4. tRNA Role: Transfer RNA (tRNA) molecules bring specific amino acids to the ribosome based on mRNA codons.
  5. Peptide Bond Formation: Amino acids are linked via peptide bonds to form a polypeptide chain.
  6. Folding & Modification: The polypeptide folds into a functional 3D protein structure, sometimes with additional modifications (e.g., glycosylation).

Key Point: The sequence of amino acids (primary structure) determines the protein’s final shape and function.

(b)(i) Decreased Reaction Rate:

Molecule D acts as a competitive inhibitor:

  • D has a shape complementary to enzyme 1’s active site, competing with substrate A for binding.
  • When D binds, it blocks A from entering the active site, reducing the breakdown of A into B and C.

Note: The inhibition is reversible; increasing substrate A concentration can overcome the effect of D.

(b)(ii) Enzyme Denaturation (Fig. 3.3):

Two environmental changes causing the distorted enzyme shape:

  1. High Temperature: Excessive heat breaks hydrogen bonds and other weak interactions, altering the active site.
  2. Extreme pH: Acids or bases disrupt ionic bonds and change the enzyme’s charge distribution, leading to misfolding.

Effect: Denaturation is irreversible—the enzyme permanently loses its catalytic function.

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