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DNA Past Paper Questions. 1. Draw as simple diagram of the molecular structure of DNA. 5 marks.

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Presentation on theme: "DNA Past Paper Questions. 1. Draw as simple diagram of the molecular structure of DNA. 5 marks."— Presentation transcript:

1 DNA Past Paper Questions

2 1. Draw as simple diagram of the molecular structure of DNA. 5 marks

3 two sugar-phosphate backbones shown A with T and C with G double helical shape shown antiparallel nature of strands indicated ten base pairs per turn of helix correct hydrogen bonding shown (A=T and C=G)

4 2. Describe the genetic code. 6 marks

5 composed of mRNA base triplets called codons 64 different codons each codes for the addition of an amino acid to a growing polypeptide chain the genetic code is degenerate meaning more than one codon can code for a partiuclar amino acid the genetic code is universal meaning it is the same in almost all organisms (AUG is the) start codon some (nonsense) codons code for the end of translation

6 3. Explain the relationship between genes and polypeptides. 5 marks

7 genes code for proteins/ polypeptides one gene one polypeptide (one) gene is transcribed into (one) mRNA mRNA is translated by a ribosome to synthesize a polypeptide if the information on a gene is changed/ mutated this may alter the structure of a protein genetic information transcribed by eukaryotes is edited before it is translated polypeptides may be altered before they become fully functional proteins

8 4. Living organisms use DNA as their genetic material. Explain how DNA is replicated within the cells of living organisms 8 marks

9 helix is unwound two strands are separated helicase (is the enzyme that unwinds the helix separating the two strands) by breaking hydrogen bonds between bases new strands formed on each of the two single strands nucleotides added to form new strands complementary base pairing A to T and G to C DNA polymerase forms the new complementary strands replication is semi-conservative each of the DNA molecules formed has one old and one new strand

10 5. Explain the significance of complementary base pairing in DNA replication 4 marks

11 when 2 DNA strands separate, each acts as a template for DNA synthesis DNA polymerase adds free nucleotides to the parent/template strand DNA polymerase adds nucleotides according to the complementary base pairing rules A=T, G=C the two double-stranded DNA molecules produced are identical to the parent DNA

12 6. Compare DNA & RNA structure: 3 marks

13 DNA RNA sugar deoxyribose ribose pyrimidinesthymine uracil strands double single

14 7. Outline DNA transcription. 7 marks

15 initiation RNA polymerase is an enzyme complex which unwinds and unzips DNA double strand RNA polymerase attaches to promoter region of gene, which marks the beginning point for transcription elongation RNA polymerase uses DNA the anti-sense strand as a template RNA polymerase synthesizes a complementary RNA strand using base pairing rules A = UT = AG = CC = G termination RNA polymerase reaches termination region of the gene, which marks the end of the coding sequence terminates transcription by releasing both DNA and RNA maximum 3 marks from each of initiation, elongation, termination

16 8. Explain the process of translation. 8 marks

17 Initiation mRNA binds to the small subunit of the ribosome tRNA anticodon binds to mRNA codon by complementary base pairing large ribosomal subunit binds, completing ribosomal structure, Elongation: tRNA with anticodon complementary to second mRNA codon binds to ribosome with appropriate amino acid attached to tRNA enzymes in ribosome catalyze formation of peptide bond between 1st and 2nd amino acids 1st tRNA (separated from amino acid) exits ribosome ribosome moves one codon (3 nucleotides) along the mRNA tRNA with anticodon complementary to next mRNA codon binds to ribosome with appropriate amino acid attached enzymes in ribosome catalyze formation of peptide bond between 2nd and 3rd amino acids repetition of process until stop codon is reached Termination: when ribosome reaches a stop codon, no tRNA has a complementary anticodon release factor protein binds to ribosome stop codon polypeptide and mRNA are released large and small ribosomal subunits separate maximum 3 marks from each of initiation, elongation, termination


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