Mendel's laws: dominance, segregation, independent assortment
Mendel grew garden peas for about 7 years. Peas were a smart choice: they grow fast, have clear opposite traits (tall/dwarf, round/wrinkled) and can self-pollinate or be crossed by hand.
Words first: a gene is a unit that controls a trait. Its different forms are alleles (T and t). Genotype = the letters (TT, Tt, tt). Phenotype = what you see (tall, dwarf). Homozygous = same letters (TT); heterozygous = different letters (Tt).
- Law of dominance: in Tt, only T shows. T is dominant, t is recessive.
- Law of segregation: the two alleles separate when gametes form. Each gamete gets only one. That is why Tt × Tt gives 1 TT : 2 Tt : 1 tt (genotype) and 3 tall : 1 dwarf (phenotype).
- Law of independent assortment: alleles of two different genes go into gametes on their own. RrYy makes four kinds of gametes (RY, Ry, rY, ry) in equal numbers, so the F2 phenotype ratio is 9 : 3 : 3 : 1.
Test cross
To find out if a tall plant is TT or Tt, cross it with a dwarf (tt). All tall → it was TT. Half tall, half dwarf (1 : 1) → it was Tt.
Incomplete dominance and co-dominance
Incomplete dominance: the heterozygote looks in between. Snapdragon (dog flower): RR red × rr white → Rr pink. F2: 1 red : 2 pink : 1 white. Here the phenotype ratio equals the genotype ratio. Mendel's idea of alleles is still right; only 'dominance' is partial.
Co-dominance: the heterozygote shows both traits fully, side by side. A person with Iᴬ and Iᴮ has both A and B sugars on red blood cells → group AB.
Simple test: mixed colour = incomplete; both shown = co-dominance.
Multiple alleles and ABO blood groups
When a gene has more than two alleles in the population, we call them multiple alleles. One person still carries only two.
The ABO gene (called I) has three alleles: Iᴬ, Iᴮ, i. Iᴬ and Iᴮ are co-dominant to each other; both are dominant over i.
| Genotype | Blood group |
|---|---|
| IᴬIᴬ, Iᴬi | A |
| IᴮIᴮ, Iᴮi | B |
| IᴬIᴮ | AB |
| ii | O |
Three alleles give 6 genotypes and 4 blood groups. Formula: n alleles → n(n+1)/2 genotypes.
Pleiotropy and polygenic traits
Pleiotropy: one gene, many effects. In phenylketonuria (PKU), one faulty gene for an enzyme leads to mental disability, light hair and light skin. In pea, one gene controls starch in seeds: it decides seed shape and also starch grain size.
Polygenic traits: many genes, one trait. Human skin colour and height are controlled by several genes, each adding a small bit. So we see a smooth range of shades, not just 2–3 types. The environment (sunlight, food) also affects them.
Memory: Pleio = plenty of effects; Poly = plenty of genes.
Chromosome theory of inheritance
Mendel's 'factors' were unknown objects in 1865. By 1900 microscopes showed chromosomes moving in meiosis. Sutton and Boveri saw the match:
- Chromosomes come in pairs, like alleles.
- The pairs separate in meiosis, like segregation.
- Different pairs line up independently, like independent assortment.
So they said: genes are on chromosomes. T.H. Morgan proved it with fruit flies (Drosophila): they breed fast, have many young, only 4 chromosome pairs, and easy-to-see traits.
Linkage and crossing over
Genes on the same chromosome tend to go together into a gamete. This is linkage. Morgan found that such genes did not give 9 : 3 : 3 : 1.
But sometimes the partner chromosomes swap pieces in meiosis (crossing over). This makes new combinations called recombinants.
- Genes close together → rarely separated → tightly linked (few recombinants, e.g. 1.3%).
- Genes far apart → often separated → loosely linked (more recombinants, e.g. 37.2%).
Recombination frequency (%) = recombinant offspring ÷ total × 100. Sturtevant used this % as a distance to build the first gene maps (1% = 1 map unit).
Sex determination in humans, birds and honeybee
Henking saw a special body in insect sperm and called it the X body. Sex chromosomes differ between males and females; the rest are autosomes.
- Humans (XX–XY): female 44 + XX makes only X eggs; male 44 + XY makes X or Y sperm (50 : 50). The sperm decides the child's sex. Male heterogamety.
- Grasshopper (XX–XO): male has only one X.
- Birds (ZW–ZZ): the female is ZW and makes two kinds of eggs; the male is ZZ. Female heterogamety — the mother decides.
- Honeybee (haplo-diploid): a fertilised egg (2n = 32) becomes a female (queen or worker). An unfertilised egg (n = 16) becomes a male drone by parthenogenesis. Drones make sperm by mitosis; they have no father and no sons, but have a grandfather.
Sex-linked traits: haemophilia and colour blindness
Some genes are on the X chromosome and the Y has no matching copy. A male has one X, so one faulty allele is enough to show the disorder. A female needs two; with one she is a carrier.
- Colour blindness (red-green): about 8% of males and 0.4% of females.
- Haemophilia: blood does not clot well; a small cut can bleed for long. Famous in Queen Victoria's family.
Pattern: carrier mother × normal father → half the sons affected, half the daughters carriers. An affected father never passes it to his son (he gives Y to sons).
Mendelian disorders: thalassemia, sickle-cell and more
A change in one gene can cause a disorder that follows Mendel's rules. A pedigree (family tree with squares = males, circles = females, filled = affected) helps trace it.
- Thalassemia: autosomal recessive. The body makes too little of one globin chain of haemoglobin, so red cells break and there is anaemia. α-thalassemia: genes HBA1 and HBA2 on chromosome 16. β-thalassemia: gene HBB on chromosome 11. It is a quantity problem (too little globin).
- Sickle-cell anaemia: autosomal recessive. One base change GAG → GUG puts valine instead of glutamic acid at the 6th place of β-globin. Red cells turn sickle-shaped at low oxygen. It is a quality problem (wrong globin).
- Haemophilia, colour blindness: X-linked recessive. Phenylketonuria: autosomal recessive.
Chromosomal disorders: Down, Turner, Klinefelter
If chromosomes fail to separate in meiosis (non-disjunction), a gamete can get one extra or one less. Aneuploidy = gain or loss of a chromosome. Polyploidy = whole extra sets (common in plants).
| Syndrome | Chromosomes | Key signs |
|---|---|---|
| Down | 47, extra chromosome 21 (trisomy 21) | short, small round head, furrowed tongue, open mouth, broad palm with a single crease, slow mental and physical growth |
| Klinefelter | 47, XXY (male) | tall, some female traits such as breast growth (gynaecomastia), sterile |
| Turner | 45, XO (female) | short, ovaries not developed, sterile, secondary sex traits missing |
Key formulas and definitions
- Monohybrid F2 (Tt × Tt): phenotype 3 : 1, genotype 1 : 2 : 1
- Test cross (Tt × tt): 1 : 1
- Dihybrid F2 (RrYy × RrYy): 9 : 3 : 3 : 1
- Incomplete dominance F2: 1 : 2 : 1 (phenotype = genotype)
- Kinds of gametes = 2ⁿ (n = heterozygous gene pairs)
- Genotypes from n alleles = n(n + 1)/2
- Recombination frequency = recombinants ÷ total offspring × 100 (1% = 1 map unit)
Worked examples
1. A pea plant Tt is selfed. Out of 400 seedlings, how many are expected to be dwarf?
Step 1: Tt × Tt → 1 TT : 2 Tt : 1 tt. Step 2: Dwarf = tt = 1/4. Step 3: 400 × 1/4 = 100 dwarf plants.
2. A tall plant is crossed with a dwarf plant. The offspring are 52 tall and 48 dwarf. What was the tall parent’s genotype?
Step 1: This is a test cross (with tt). Step 2: About 1 : 1 means the tall parent gave T and t equally. Answer: Tt.
3. How many kinds of gametes does AaBbCC make?
Step 1: Count heterozygous pairs: Aa and Bb = 2 (CC is homozygous). Step 2: 2ⁿ = 2² = 4 kinds: ABC, AbC, aBC, abC.
4. In a dihybrid cross RrYy × RrYy, 1600 seeds are formed. How many are round and green?
Step 1: F2 ratio 9 RY : 3 Ry : 3 rY : 1 ry. Step 2: Round green = R_yy = 3/16. Step 3: 1600 × 3/16 = 300 seeds.
5. A mother is blood group A (Iᴬi) and the father is B (Iᴮi). What fraction of children can be O?
Step 1: Gametes: mother Iᴬ or i; father Iᴮ or i. Step 2: Boxes: IᴬIᴮ (AB), Iᴬi (A), Iᴮi (B), ii (O). Step 3: O = 1/4 = 25%.
6. Pink snapdragons are crossed with each other and give 240 plants. How many are pink?
Step 1: Rr × Rr → 1 RR : 2 Rr : 1 rr. Step 2: Pink = Rr = 2/4. Step 3: 240 × 1/2 = 120 pink.
7. A test cross of a dihybrid gives 420 parental-type and 80 recombinant offspring. Find the recombination frequency and map distance.
Step 1: Total = 420 + 80 = 500. Step 2: RF = 80/500 × 100 = 16%. Step 3: Map distance = 16 map units (cM). The genes are linked, but not tightly.
8. A colour-blind man marries a woman with normal vision who is not a carrier. What are the chances their son is colour-blind? Their daughter?
Step 1: Man XᶜY, woman XX. Step 2: Sons get Y from father and X from mother → XY, all normal (0%). Step 3: Daughters get Xᶜ from father and X from mother → XᶜX, all carriers, none affected.
Common mistakes
- Mixing up incomplete dominance and co-dominance. Pink (a blend) is incomplete; AB blood (both shown) is co-dominance.
- Thinking one person can carry all three blood alleles. The population has three; one person has only two.
- Saying the mother decides the sex of the baby in humans. The father’s X or Y sperm decides. (In birds it is the mother.)
- Writing 9 : 3 : 3 : 1 for linked genes. Linked genes give mostly parental types, so the ratio changes.