The roles of genes in determining the phenotype

Biology A2 · Inheritance. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A gene is a length of DNA that codes for a polypeptide. It sits at a fixed position, its locus, on a chromosome. Different forms of a gene are alleles. The alleles an organism has are its genotype; the features that result are its phenotype.

A genetic diagram follows alleles from parents into gametes and then into offspring, and predicts the ratio of phenotypes. The ratio changes with dominance, codominance, multiple alleles, sex linkage, autosomal linkage and epistasis.

Real results never match the prediction exactly. The chi-squared test shows whether the difference between observed and expected numbers is small enough to be due to chance.

Rules to remember

Monohybrid ratios: heterozygote × heterozygote gives 3 : 1 (1 : 2 : 1 with codominance); a test cross (× homozygous recessive) of a heterozygote gives 1 : 1
Dihybrid ratios for unlinked genes: double heterozygote × double heterozygote gives 9 : 3 : 3 : 1; its test cross gives 1 : 1 : 1 : 1
Sex linkage: the gene is on the X chromosome, so a male has only one allele and shows it; a father passes his X to all his daughters and to none of his sons
Autosomal linkage: genes on the same chromosome are inherited together, so a test cross gives many parental types and few recombinants (from crossing over)
Epistasis: one gene affects the expression of another gene at a different locus, which changes the 9 : 3 : 3 : 1 ratio
χ² = Σ (O − E)² ÷ E; degrees of freedom = number of classes − 1; if χ² is less than the critical value at p = 0.05, the difference is not significant
TYR → tyrosinase (none: albinism); HBB → β-globin (sickle cell anaemia); F8 → factor VIII (haemophilia, sex-linked); HTT → huntingtin (Huntington's disease, dominant); Le → enzyme that makes active gibberellin (tall pea plants)

Common mistake

In the chi-squared test, use the actual numbers counted, not ratios or percentages. Remember that a value below the critical value means the results fit the expected ratio.

Worked example

A cross expected to give a 3 : 1 ratio produces 84 tall and 36 dwarf plants. Calculate χ². Is the difference significant? (Critical value at p = 0.05 for 1 degree of freedom is 3.84.)

Total = 84 + 36 = 120, so expected numbers are 120 × 3/4 = 90 tall and 120 × 1/4 = 30 dwarf.
Tall: (84 − 90)² ÷ 90 = 36 ÷ 90 = 0.4. Dwarf: (36 − 30)² ÷ 30 = 36 ÷ 30 = 1.2.
χ² = 0.4 + 1.2 = 1.6, which is less than 3.84.

Answer: χ² = 1.6; the difference is not significant and the results fit a 3 : 1 ratio.

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