D3.2 InheritanceIB Biology HL: Revision notes
Section 1
Gametes, crosses, genotype and phenotype
Parents make haploid gametes that fuse to form a diploid zygote, so a diploid cell has two copies of each autosomal gene. In flowering plants, pollen carries male gametes and female gametes are in the ovary; peas can self-pollinate, so anthers are removed before a controlled cross. Terms: P, F1, F2 generation, Punnett grid. Crosses are used to breed new crop and ornamental varieties.
The genotype is the alleles inherited (homozygous or heterozygous); the phenotype results from genotype and environment. A dominant allele gives the same phenotype when homozygous or heterozygous because one copy produces enough functional protein. Phenotypic plasticity is developing traits suited to the environment by changing gene expression, without changing genotype, and may be reversible.
Section 2
PKU, multiple alleles, codominance and incomplete dominance
PKU is caused by a recessive allele of an autosomal gene for the enzyme converting phenylalanine to tyrosine. New alleles often arise as SNPs; any number of alleles can exist in a gene pool but an individual has two. ABO blood groups have three alleles: , , .
Codominance: heterozygote has a dual phenotype (, group AB). Incomplete dominance: heterozygote is intermediate (pink Mirabilis jalapa from red × white; F2 1 red : 2 pink : 1 white).
Section 3
Sex linkage, pedigrees and continuous variation
Females are XX, males XY; the sperm determines sex. The X carries many more genes than the Y, so males are affected by a single recessive X-linked allele such as the haemophilia allele ; females are usually carriers. Pedigree charts reveal patterns: unaffected parents with an affected child indicate a recessive allele. Deducing a pattern from cases is inductive; applying it to an individual is deductive. Marriage between close relatives is prohibited in many societies because relatives are likely to share recessive alleles.
Continuous variation (height, skin colour) arises from polygenic inheritance and environment; discrete variables (ABO group) fall into classes. A box-and-whisker plot shows minimum, Q1, median, Q3, maximum and outliers (more than 1.5 × IQR beyond a quartile).
Section 4
HL: Segregation, independent assortment and dihybrid crosses
In meiosis, segregation separates the two alleles of each gene into different gametes. For unlinked genes (on different chromosomes), the random orientation of bivalents at metaphase I means the alleles of one gene assort independently of the other.
- AaBb produces AB, Ab, aB and ab gametes in equal proportions.
- AaBb × AaBb: a 4 × 4 Punnett grid gives 9 : 3 : 3 : 1.
- AaBb × aabb (test cross): 1 : 1 : 1 : 1.
NOS: this is Mendel's second law. It only applies if genes are on different chromosomes or far enough apart on one chromosome for recombination to reach 50%; biological "laws" have exceptions.
Always list the four gamete types first, then fill the grid. Most dihybrid errors come from wrong gametes.
Section 5
HL: Gene loci, linkage and recombinants
Each gene has a locus on a chromosome, e.g. PAH at 12q23.2, and codes for a polypeptide; databases list loci, some on different chromosomes and some close together on one chromosome. Linked genes are on the same chromosome, so their alleles tend to be inherited together and fail to assort independently. Show linked genotypes with alleles beside vertical lines for the homologous chromosomes, e.g. GN / gn.
Crossing over at a chiasma in prophase I creates recombinant chromosomes. In a test cross of a double heterozygote with a double homozygous recessive, the two most frequent offspring classes are the parental types and the two rare classes are recombinants. Recombinants can be identified in gametes, offspring genotypes and offspring phenotypes. Recombination frequency = recombinants ÷ total; it is below 50% for linked genes.
Unlinked genes also give recombinants (half the test-cross offspring). Linkage is shown by recombinants being much fewer than 50%.
Section 6
HL: Chi-squared test on dihybrid data
The chi-squared test asks whether the difference between observed and expected counts is due to chance.
- Null hypothesis: no significant difference from the expected ratio (e.g. 9 : 3 : 3 : 1); alternative: there is a difference.
- Expected = total × ratio fraction.
- .
- Degrees of freedom = number of classes − 1.
- Compare with the critical value at p = 0.05. If χ² is smaller, accept the null hypothesis; if larger, reject it (e.g. evidence of linkage).
NOS: the F2 is a sample used to represent the population of all possible offspring.
That's the notes covered.
Carry on to the next subtopic.