Genetic Information, Variation and Relationships
Genetic Information, Variation and Relationships is the A-level Biology topic covering DNA structure and replication, the genetic code, transcription and translation, gene mutations, meiosis, and biodiversity and classification. It links the molecular biology of earlier topics to how genetic variation arises and is measured. Expect base-counting, amino-acid-coding and biodiversity-index calculations.
Method
- Learn the structure of a DNA nucleotide (phosphate, deoxyribose sugar, nitrogenous base) and the complementary base pairing rule (A-T, C-G) before tackling any base-counting question.
- Work through semi-conservative replication and transcription/translation as a sequence of named enzymes and steps, rather than memorising a single paragraph: helicase unwinds, DNA polymerase builds new strands, RNA polymerase builds pre-mRNA, splicing removes introns.
- For coding calculations, remember: number of codons = number of bases / 3, and the number of amino acids in the finished polypeptide is one fewer than the number of codons if a stop codon is included.
- Classify a mutation (substitution, insertion or deletion) before explaining its effect: a deletion or insertion usually causes a frameshift affecting every codon downstream, whereas a substitution affects at most one codon.
- For meiosis and variation questions, separate the three sources of variation clearly: independent assortment, crossing over, and random fertilisation.
- For biodiversity index calculations (e.g. Simpson's Index), write out each species' n/N value before squaring and summing, to avoid arithmetic slips.
Worked example
A section of DNA contains 4,600,000 bases in total, of which 28% are adenine. Calculate the number of cytosine bases in this section of DNA.
- Because A pairs with T, the percentage of thymine also equals 28% (%A = %T).
- The remaining bases are cytosine and guanine, which are also equal to each other (%C = %G): 100 - (28 + 28) = 44%, so %C = %G = 44 / 2 = 22%.
- Calculate the number of cytosine bases: 22% of 4,600,000 = 0.22 x 4,600,000.
- = 1,012,000.
- Final answer: 1,012,000 cytosine bases.
Practice questions
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Q1A single DNA nucleotide is built from three different components. Name all three.Show answer
Answer: A phosphate group, a deoxyribose (pentose) sugar, and a nitrogen-containing base.
Q2Name the enzyme that joins adjacent nucleotides together during DNA replication.Show answer
Answer: DNA polymerase.
Q3State the type of mutation in which a single DNA base is deleted, and state the effect this typically has on the reading frame.Show answer
Answer: A deletion mutation; it causes a frameshift, changing every codon (and usually every amino acid) downstream of the mutation.
Q4State the number of chromosomes present in a human gamete.Show answer
Answer: 23
Q5A gene with no introns has a coding sequence of 891 base pairs, and the finished polypeptide includes a stop codon. Calculate the number of amino acids in the polypeptide.Show answer
Answer: 296 (891 / 3 = 297 codons; 297 - 1 for the stop codon = 296 amino acids)
Q6A rocky shore survey recorded four species with the following counts: species A = 30, species B = 45, species C = 15, species D = 10 (total N = 100). Calculate Simpson's Index of Diversity, D = 1 - sum(n/N)^2, to 3 significant figures.Show answer
Answer: 0.675 (sum of (n/N)^2 = 0.09 + 0.2025 + 0.0225 + 0.01 = 0.325; D = 1 - 0.325 = 0.675)
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
Identify three properties that describe the nature of the genetic code.
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Meselson and Stahl grew bacteria in a medium containing only heavy nitrogen (15N), so all their DNA became heavy. The bacteria were then transferred to a medium containing only light nitrogen (14N) and allowed to complete exactly two rounds of DNA replication. Predict, with reasons, the density (or densities) of DNA that would be found after these two rounds of replication if DNA replication is semi-conservative.
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A gene contains a coding sequence of 1,203 base pairs with no introns. (a) Calculate the maximum number of amino acids that could be coded for by this gene, excluding the stop codon. (b) A single base is substituted partway through the gene, changing only one codon, but the amino acid coded for does not change. State and explain the term used to describe this type of mutation.
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