Module 3: Genetics - DNA, Gene Expression, Meiosis and Inheritance
Module 3 covers the storage, replication and expression of genetic information, the behaviour of chromosomes in meiosis, and the inheritance patterns that follow. Expect questions that ask you to trace information flow and predict ratios from crosses.
DNA structure and replication
Nucleotides: deoxyribose, phosphate, and a base - purines adenine and guanine, pyrimidines thymine and cytosine. Nucleotides link 3′–5′ by phosphodiester bonds into two antiparallel strands; A pairs with T by two hydrogen bonds, G with C by three. Replication is semi-conservative: helicase unwinds, each strand acts as a template, DNA polymerase adds nucleotides 5′→3′, the leading strand is continuous while the lagging strand forms Okazaki fragments joined by ligase. Meselson–Stahl's density-gradient experiment supports this.
Gene expression
The code is a triplet, degenerate, non-overlapping and (nearly) universal, with AUG as start and UAA/UAG/UGA as stops. Transcription: RNA polymerase makes mRNA from the template strand; eukaryotic mRNA is spliced to remove introns. Translation: mRNA binds a ribosome, tRNA anticodons pair with codons, peptide bonds form between amino acids, and the polypeptide is released at a stop codon.
Meiosis and inheritance
Meiosis I separates homologous chromosomes, meiosis II separates sister chromatids, giving four haploid cells. Crossing over at chiasmata in prophase I and independent assortment in metaphase I generate variation. Monohybrid crosses give 3:1 in the F2, or 1:1 in a test cross with a homozygous recessive. Codominance and multiple alleles (ABO blood groups) break simple dominance; sex-linked genes give different ratios in males and females because males are hemizygous. Dihybrid crosses give 9:3:3:1 only if the genes are unlinked; linked genes give excess parental types, with recombinants arising only from crossing over.
Worked example
A grey, normal-winged fly is test-crossed with an ebony, vestigial fly and gives 41 grey normal, 39 ebony vestigial, 10 grey vestigial, 9 ebony normal. Equal numbers of the four classes would mean unlinked genes (1:1:1:1). Here parentals (80) far exceed recombinants (19), so the genes are linked on the same chromosome. Recombinant frequency = 19/99 ≈ 19%, which is the map distance in centimorgans, so the loci are about 19 cM apart. Reading the ratio as 9:3:3:1, or assuming independent assortment, would miss this.
Traps that cost marks
Confusing the template strand with the coding strand. Saying DNA polymerase works 3′→5′. Giving genotype ratios where gamete or phenotype ratios are asked. Using a test cross with anything other than a homozygous recessive. Assuming independent assortment for linked genes. Forgetting that crossing over occurs in prophase I, not anaphase.