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Questions on the genetic code and transcription
The genetic code as triplet, degenerate, non-overlapping and near-universal, with the consequence of each property; genes, introns and exons; transcription by RNA polymerase step by step; the splicing of pre-mRNA into mature mRNA; and why degeneracy makes many base substitutions silent.
15 original questions · 49 marks · the genetic code and transcription notes · Nucleic acids, genomes and protein synthesis
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Describe how a molecule of pre-mRNA is produced from a gene in the nucleus of a eukaryotic cell.
Mark scheme
- B1 RNA polymerase binds to a promoter region at the start of the gene
- B1 the enzyme unwinds a short section of the double helix, breaking the hydrogen bonds between the paired bases
- B1 only one of the two exposed strands, the template strand, is read
- B1 free RNA nucleotides align against the template by complementary base pairing, with uracil pairing to adenine wherever thymine would have gone
- B1 RNA polymerase joins the aligned nucleotides by condensation into phosphodiester bonds, moving along the gene until it reaches a terminator sequence and releases the pre-mRNA
Explain why a mature mRNA molecule is shorter than the gene it was transcribed from, and explain how one gene can give rise to more than one polypeptide.
Mark scheme
- B1 the pre-mRNA is a copy of the whole gene, introns included
- B1 spliceosomes cut the introns out and join the exons together, so the introns are not represented in the mature mRNA
- B1 in alternative splicing, different combinations of exons are retained in the mature mRNA
- B1 so several different mRNA molecules, and therefore several different polypeptides, can come from the same gene
A single base is inserted into the coding sequence of a gene close to its start. The polypeptide produced is much shorter than normal and its amino acid sequence bears no relation to the original beyond the point of insertion. Suggest why, referring to two properties of the genetic code.
Mark scheme
- B1 the code is read in threes from a fixed starting point and is non-overlapping, so each base belongs to one triplet only
- B1 inserting a base pushes every subsequent base one place along, so every triplet after the insertion is read differently, which is a frame shift
- B1 the amino acid sequence after that point is therefore completely altered rather than altered in one place
- B1 a shifted reading frame produces stop codons at positions where there were none, so translation ends early and the polypeptide is short
Compare the coding strand of a gene with the template strand, referring to which one is transcribed and how each one's sequence relates to the mRNA produced.
Mark scheme
- B1 the template strand is the strand that RNA polymerase actually reads and copies, whereas the coding strand is not transcribed at all
- B1 the mRNA produced is complementary to the template strand, base for base, whereas the mRNA has the same sequence as the coding strand except that every thymine in the coding strand is replaced by uracil in the mRNA
- B1 given the coding strand, the mRNA sequence can be written directly by substituting U for T, whereas given the template strand, the complementary sequence must first be taken before substituting U for T
- B1 only the template strand is exposed to complementary base pairing with the incoming RNA nucleotides during transcription, so only its sequence, not the coding strand's, physically determines which bases are added
A textbook describes introns and other non-coding DNA as 'junk DNA' with no function. Evaluate this description, using what you know about non-coding DNA.
Mark scheme
- B1 for: introns and much other non-coding DNA does not code for a polypeptide, and it is removed from pre-mRNA by splicing before the mature mRNA is used, so it plays no part in specifying the sequence of the finished protein
- B1 against: a large proportion of eukaryotic DNA is non-coding, including introns, repeated sequences between genes and regions involved in switching genes on and off, and several of these regions have known regulatory or other roles rather than none at all
- B1 against: alternative splicing depends on retaining different combinations of exons from the same pre-mRNA, so which sequence is treated as exon and which as intron contributes to which mRNA molecules, and hence which polypeptides, a gene can produce
- B1 judgement: 'non-coding' is not the same claim as 'functionless', and describing all such DNA uniformly as junk overstates how much of it has been shown to do nothing, even though a function for every stretch of it has not been demonstrated either
A student says that the genetic code being degenerate, with most amino acids specified by more than one triplet, shows it is a wasteful system that evolution has simply not got round to tidying up. Discuss this claim.
Mark scheme
- B1 for the student's claim: sixty-one triplets are shared between only twenty amino acids, which is forty-one more than the minimum needed, so in terms of the sheer number of triplets in use there is apparent redundancy
- B1 against: triplets sharing an amino acid tend to differ mainly in their third base, so a substitution mutation at that position is often silent, leaving the protein completely unchanged rather than harming it
- B1 against: because degeneracy makes many mutations silent, it reduces how often a random copying error or mutagen actually changes the protein an organism produces, which is a protective effect rather than a wasted one
- B1 judgement: describing the redundancy as unused waste ignores this protective consequence; degeneracy is better understood as a feature that has been retained because it reduces the impact of mutation, not as an oversight evolution failed to remove
The coding strand of a short section of a gene reads GAT CCA TTG. Give the base sequence of the template strand, written base for base beneath the coding strand, give the base sequence of the mRNA transcribed from this section, and give the rule that relates the mRNA sequence to the coding strand.
Mark scheme
- B1 template strand CTA GGT AAC, each base complementary to the coding strand
- A1 mRNA GAU CCA UUG
- B1 the mRNA matches the coding strand, with uracil in place of every thymine
A base substitution changes the third base of one mRNA codon, and the polypeptide produced is completely unchanged. Explain how this is possible.
Mark scheme
- B1 the genetic code is degenerate: most amino acids are specified by more than one triplet
- B1 triplets sharing an amino acid tend to differ in the third base, so the new codon can specify the same amino acid as the old one
- B1 the primary structure is then identical, so the polypeptide is unchanged, and the substitution is called silent
State what is meant by a gene, state what is meant by its locus, and state how the genome differs from the proteome.
Mark scheme
- B1 a gene is a sequence of DNA bases that codes for a polypeptide, or for a functional RNA molecule such as a tRNA
- B1 the locus of a gene is its specific position on a chromosome
- B1 the genome is the complete set of DNA in a cell, whereas the proteome is the full range of proteins that cell is able to produce
Describe the difference between an exon and an intron, and describe what happens to each during splicing.
Mark scheme
- B1 an exon is a section of a gene that ends up represented in the mature mRNA, whereas an intron is a non-coding section that does not
- B1 the pre-mRNA produced by transcription contains both exons and introns, since it is a copy of the whole gene
- B1 spliceosomes cut the introns out of the pre-mRNA and join the exons together, producing the mature mRNA that leaves the nucleus
Explain why the terminator sequence at the end of a gene and a stop codon in an mRNA molecule are two different things, even though both are commonly called 'stops'.
Mark scheme
- B1 the terminator is a sequence in the DNA of the gene itself, recognised by RNA polymerase, which causes the enzyme to detach and release the finished pre-mRNA, ending transcription
- B1 a stop codon is a triplet in the mature mRNA, read by the ribosome during translation, which causes translation of the polypeptide to end
- B1 the terminator therefore acts on RNA polymerase and finishes making the RNA copy, while a stop codon acts on the ribosome and finishes making the polypeptide
Mitochondrial DNA interprets a small number of triplets differently from nuclear DNA in the same cell. Suggest what this shows about the claim that the genetic code is universal, and suggest why the near-universality of the code across most organisms is still significant for genetic engineering.
Mark scheme
- B1 the code is not absolutely universal, since mitochondria interpret a few triplets differently from the interpretation used by the nuclear genome of the very same cell, so 'universal' should be read as 'almost universal'
- B1 despite that exception, the code is read the same way across the great majority of triplets and the great majority of organisms, including between organisms as different as bacteria and humans
- B1 that near-universal agreement is what allows a gene taken from one species to be transcribed and translated correctly by the machinery of an unrelated species, which is what genetic engineering that inserts a human gene into a bacterium depends on
State what is meant by saying that the genetic code is degenerate, and state what is meant by saying that it is non-overlapping.
Mark scheme
- B1 degenerate: most amino acids are specified by more than one triplet
- B1 non-overlapping: each base belongs to one triplet only, and the triplets are read one after another
DNA uses four bases read in triplets. Calculate the total number of different triplets, and the number left to specify amino acids once the stop signals are set aside.
Mark scheme
- M1 the number of triplets is 4 raised to the power 3, which is 64
- A1 three of the 64 are stop signals, leaving 61 that specify amino acids
Name the enzyme that carries out transcription, and name the structures that remove introns from pre-mRNA.
Mark scheme
- B1 RNA polymerase, which is not the enzyme that replicates DNA
- B1 spliceosomes, which cut the introns out and join the exons together
Practise the genetic code and transcription one question at a time
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