DNA, genes and protein synthesisAQA A-Level Biology: Topic test

1. 1
   Researchers isolated DNA from the nuclei and from the mitochondria of mouse liver cells and examined its structure.

   (a)

   Which row correctly describes the mitochondrial DNA?

   [1 mark]

   • ALinear and associated with histones
   • BShort, circular and not associated with proteins
   • CVery long, linear and containing many introns
   • DEnclosed by a nuclear envelope

   (b)

   Which row correctly describes the mouse nuclear DNA?

   [1 mark]

   • AShort, circular and not associated with proteins
   • BShort, circular and associated with histones
   • CVery long, linear and associated with histones
   • DVery long, circular and associated with histones

   (c)

   Describe how a molecule of nuclear DNA and its associated proteins form a chromosome.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   A gene for green fluorescent protein from a jellyfish was inserted into the DNA of mouse cells. The mouse cells made fluorescent protein using their own ribosomes and transfer RNA. The jellyfish polypeptide has 238 amino acids.

   (a)

   What is the minimum number of mRNA bases that code for the amino acids of this polypeptide, ignoring the stop codon?

   [1 mark]

   • A714
   • B238
   • C476
   • D952

   (b)

   Several different codons can code for the amino acid leucine. This shows that the genetic code is

   [1 mark]

   • Auniversal
   • Bnon-overlapping
   • Ca doublet code
   • Ddegenerate

   (c)

   Explain how the production of fluorescent protein in mouse cells shows that the genetic code is universal.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A DNA template strand has the base sequence TAC AGG CTT GAC ATT. Part of the genetic code for the mRNA codons is: AUG = methionine (start); UCC = serine; GAA and GAG = glutamic acid; CUG = leucine; UAA = stop; CCG = proline; AAC = asparagine; UGU = cysteine.

   (a)

   Deduce the base sequence of the mRNA transcribed from this template and the amino acid sequence of the polypeptide formed.

   [3 marks]

   (b)

   A substitution changes the last base of the third template triplet from T to C (CTT to CTC). Separately, a deletion removes the first base (A) of the second template triplet. Explain the effect of each change on the polypeptide.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A hen's oviduct cell makes the protein ovalbumin. The ovalbumin gene is 7700 base pairs long and contains seven introns. The mature mRNA that is translated at the ribosomes is 1158 nucleotides long, and its final triplet is a stop codon.

   (a)

   Describe how the ovalbumin gene is used to make the mature mRNA, and explain why the mature mRNA is much shorter than the gene.

   [6 marks]

   (b)

   Calculate the number of amino acids in the ovalbumin polypeptide, and describe how the polypeptide is assembled from the mature mRNA at a ribosome.

   [6 marks]

   Total for question 4: 12 marks

5. 5
   An antibiotic called rifampicin binds to RNA polymerase in bacteria and stops it joining nucleotides together.

   (a)

   Which process in a bacterium is directly prevented by the antibiotic?

   [1 mark]

   • ATranslation of mRNA at ribosomes
   • BReplication of the bacterial DNA
   • CAttachment of amino acids to tRNA
   • DTranscription of genes into mRNA

   (b)

   Why does protein synthesis in treated bacteria eventually fall to a very low level?

   [1 mark]

   • ANo new mRNA is made to be translated
   • BPeptide bonds can no longer form between amino acids
   • CThe genetic code becomes degenerate
   • DIntrons can no longer be removed from pre-mRNA

   (c)

   Explain why mRNA made by a bacterium can be translated immediately, whereas the product of transcription in a human cell cannot.

   [2 marks]

   Total for question 5: 4 marks

6. 6
   A student compares a bacterium that has a genome of 4300 genes with a parasitic bacterium that has only 480 genes.

   (a)

   What is the proteome of a cell?

   [1 mark]

   • AThe complete set of genes in the cell
   • BThe full range of proteins that the cell is able to produce
   • CAll of the DNA in the nucleus
   • DAll of the mRNA molecules in the cytoplasm

   (b)

   Which statement about genes in a genome is correct?

   [1 mark]

   • AEvery gene codes for a polypeptide
   • BA gene has no fixed position on a DNA molecule
   • CA gene can code for a functional RNA such as tRNA or rRNA
   • DA gene is made only of exons in all cells

   (c)

   Suggest why the proteome of the parasitic bacterium is likely to be smaller than that of the other bacterium.

   [2 marks]

   Total for question 6: 4 marks

7. 7
   A eukaryotic chromosome consists of 60 million base pairs. Exons make up 2.4 million of these base pairs. One gene on the chromosome has 5103 bases, of which 3900 are in introns. The last three bases of the final exon are a stop codon.

   (a)

   Calculate the percentage of the chromosome's DNA that is in exons, and name the type of DNA that makes up most of the remainder.

   [3 marks]

   (b)

   Calculate the number of amino acids in the polypeptide coded for by this gene. Explain each step of your calculation.

   [4 marks]

   Total for question 7: 7 marks

8. 8
   In a cell-free system containing ribosomes, tRNA, amino acids, ATP and enzymes, a synthetic mRNA with the repeating sequence CUCUCUCUCUCU… produced a polypeptide of alternating leucine and serine. A second synthetic mRNA with the repeating sequence UUCUUCUUCUUC… produced three different polypeptides, each containing only one type of amino acid: polyphenylalanine, polyserine and polyleucine. The codons CUC and CUU both code for leucine.

   (a)

   Explain how these results show that the genetic code is a non-overlapping triplet code.

   [6 marks]

   (b)

   Explain what the results for CUC and CUU show about the genetic code, and describe how the correct amino acid is brought to each codon on the ribosome and joined to the growing polypeptide in this system.

   [6 marks]

   Total for question 8: 12 marks

End of questions