5.1 Mendel's laws
Different versions of a gene are called alleles. An individual has two alleles of each gene: if they are the same it is homozygous (AA or aa), if different heterozygous (Aa). A dominant allele (A) masks a recessive one (a); the genetic make-up is the genotype and the visible trait the phenotype. By the law of segregation, an individual's two alleles separate when gametes form, so each gamete gets only one.
In an Aa × Aa cross, the offspring genotype ratio is 1 AA : 2 Aa : 1 aa and the phenotype ratio 3 dominant : 1 recessive. By the law of independent assortment, genes on different chromosomes are sorted independently; an AaBb × AaBb cross gives a 9 : 3 : 3 : 1 phenotype ratio. In probability problems, solving each gene separately and multiplying the results is much faster than drawing large Punnett squares.
FormulaProduct rule: P(A and B) = P(A) × P(B) · Sum rule: P(A or B) = P(A) + P(B) (mutually exclusive events)
5.2 Beyond Mendel
Not every trait shows simple dominance. In codominance, both alleles show in the phenotype: in the ABO blood groups, Iᴬ and Iᴮ are codominant and i is recessive to both. Because this gene has three alleles (multiple alleles), there are four blood groups: A (IᴬIᴬ, Iᴬi), B (IᴮIᴮ, Iᴮi), AB (IᴬIᴮ) and O (ii).
Genes carried on the X chromosome show sex-linked inheritance. Males have only one X, so a recessive allele on it shows directly in the phenotype; that is why red–green colour blindness and haemophilia are much more common in males. A carrier mother (XᴬXᵃ) passes the allele to half her sons.
5.3 Linked genes and gene maps
Genes close together on the same chromosome are usually inherited together (linkage), so the 9 : 3 : 3 : 1 ratio doesn't appear. But crossing over in meiosis sometimes separates them, producing recombinant gametes. The farther apart two genes are, the more likely a crossover is between them.
The proportion of recombinant offspring gives the recombination frequency; a frequency of 1% counts as 1 map unit (centimorgan, cM). If the pairwise distances between three genes are known, their order on the chromosome can be found: the two farthest apart are at the ends and the third lies between them.
FormulaRecombination frequency (%) = recombinant offspring ÷ total offspring × 100 = map distance (cM)
Key terms
- Allele
- One of the different versions of a gene.
- Test cross
- Crossing a dominant-phenotype individual with a homozygous recessive to find its genotype.
- Codominance
- Both alleles showing in the phenotype together.
- Centimorgan (cM)
- A unit of gene distance based on recombination frequency (1% = 1 cM).
End-of-unit problems
Solve each one yourself first, then open the solution.
Two AaBbCc individuals are crossed (the genes assort independently). What fraction of the offspring will be (a) aabbcc, (b) dominant for all three traits?
Show the solution
For each gene Aa × Aa: P(aa) = 1/4, P(dominant) = 3/4. (a) 1/4 × 1/4 × 1/4 = 1/64. (b) 3/4 × 3/4 × 3/4 = 27/64.
A mother with blood group A and a father with blood group B have a child with blood group O. Find the parents' genotypes. What is the probability that their next child is AB?
Show the solution
A group-O child is ii, with one i from each parent. So the mother is Iᴬi and the father Iᴮi. In Iᴬi × Iᴮi, A, B, AB and O each occur with probability 1/4. The chance of AB is 1/4.
A woman who carries colour blindness marries a man with normal colour vision. Find the probability that a child is (a) a colour-blind son, (b) a colour-blind daughter.
Show the solution
XᴬXᵃ × XᴬY. Offspring: XᴬXᴬ, XᴬXᵃ, XᴬY, XᵃY, each 1/4. (a) A colour-blind son (XᵃY) = 1/4 of all children, or 1/2 of sons. (b) Every daughter gets Xᴬ from her father, so the probability is 0; half the daughters are carriers.
In a test cross, 415 + 405 of 1000 offspring are parental types and 92 + 88 are recombinants. What is the distance between the two genes in cM? What ratio would you expect if the genes assorted independently?
Show the solution
Recombinants: 92 + 88 = 180. Frequency = 180 ÷ 1000 × 100 = 18%, i.e. 18 cM. Independently assorting genes would give a 1 : 1 : 1 : 1 test-cross ratio (about 250 each, 50% recombination); the excess of parental types shows that the genes are linked.
🦉 An easier explanation: Professor Owl's lesson: Heredity and DNA