A-Level Biology DNA, protein synthesis and genetics: from base sequence to inheritance

Where DNA and genetics sit in each specification

The key ideas

DNA is a double helix of two antiparallel polynucleotide strands joined by hydrogen bonds between complementary bases: A with T (two hydrogen bonds) and C with G (three). Replication is semi-conservative: each new molecule keeps one original strand. DNA polymerase adds nucleotides only in the 5' to 3' direction, which is why one strand is copied continuously and the other in fragments joined by ligase.

In transcription, RNA polymerase uses the template strand to build mRNA, with U instead of T. In eukaryotes, introns are removed and exons joined. In translation, ribosomes read the mRNA codons, and tRNA molecules with complementary anticodons bring the matching amino acids, which are joined by peptide bonds. The code is a triplet code, non-overlapping and degenerate (most amino acids have more than one codon).

In a genetic cross, write the parental phenotypes and genotypes, the gametes, a Punnett square, and the offspring genotypes, phenotypes and ratio. A monohybrid cross of two heterozygotes gives 3:1; a dihybrid cross of two double heterozygotes with unlinked genes gives 9:3:3:1. Sex-linked genes are on the X chromosome, so males (XY) show recessive X-linked conditions with a single allele.

Worked example 1: from template strand to polypeptide

1. The template strand reads 3' TAC GGA TTT ACT 5'.
2. Transcription gives the complementary mRNA, with U for T: 5' AUG CCU AAA UGA 3'.
3. Read the codons from the start codon: AUG = methionine, CCU = proline, AAA = lysine, UGA = stop.
4. The polypeptide is Met-Pro-Lys.
5. If a substitution changes TTT to TTC in the template, the codon becomes AAG, which still codes for lysine: a silent mutation, because the code is degenerate. A deletion of one base would shift the reading frame and change every codon after it.

Worked example 2 (Cambridge topic 16): a dihybrid cross and chi-squared

1. Two plants heterozygous for two unlinked genes (AaBb x AaBb) produce 160 offspring. Expected ratio 9:3:3:1 gives expected numbers 90, 30, 30 and 10.
2. Observed numbers are 95, 28, 30 and 7.
3. Chi-squared = sum of (O - E)^2/E = 25/90 + 4/30 + 0/30 + 9/10 = 0.278 + 0.133 + 0 + 0.900 = 1.31.
4. Degrees of freedom = number of classes - 1 = 3. The critical value at p = 0.05 is 7.815.
5. 1.31 is less than 7.815, so the difference between observed and expected is not significant. The results are consistent with a 9:3:3:1 ratio and independent assortment.

Worked example 3: sex linkage (red-green colour blindness)

1. A carrier mother (X^B X^b) and a father with normal vision (X^B Y).
2. Gametes: mother X^B or X^b; father X^B or Y.
3. Offspring: X^B X^B (normal daughter), X^B X^b (carrier daughter), X^B Y (normal son), X^b Y (colour-blind son).
4. Each son has a 1 in 2 chance of being colour-blind; no daughter is colour-blind, but each has a 1 in 2 chance of being a carrier.

Common mistakes that cost marks

• Saying the mRNA is copied from 'the DNA' without naming the template strand, or writing T in mRNA.
• Confusing codons (on mRNA) with anticodons (on tRNA).
• Leaving gametes out of genetic diagrams, or not circling them.
• Giving a ratio when the question asks for a probability, or the reverse.
• Using degrees of freedom equal to the number of classes rather than one less.
• Concluding that chi-squared 'proves' a hypothesis rather than that the difference is or is not significant.

Exam technique and how a tutor helps

Genetics questions reward a fixed layout. Define your symbols, show parental genotypes, gametes, a Punnett square and the outcome. In protein synthesis answers, use the named molecules from the specification: RNA polymerase, mRNA, codon, tRNA, anticodon, ribosome, peptide bond. For chi-squared, state the null hypothesis, show the calculation, give degrees of freedom and the critical value, and conclude.

In one-to-one lessons a tutor works through crosses of increasing difficulty on the shared whiteboard, from monohybrid to codominance, sex linkage, dihybrid and, for Cambridge, linkage and epistasis, and checks each diagram for the small omissions that cost marks.

Self-check: can you do these?

• What mRNA codon is made from the template triplet CGT? (Answer: GCA)
• Why is replication called semi-conservative? (Answer: each new DNA molecule has one original and one new strand)
• What ratio do you expect from a test cross of Aa x aa? (Answer: 1:1)
• Why are X-linked recessive conditions more common in males? (Answer: males have one X chromosome, so one recessive allele is expressed)
• Chi-squared is 9.2 with 3 degrees of freedom. Is it significant at p = 0.05? (Answer: yes, 9.2 > 7.815)