Zea mays is a monoecious plant with
(i) Male inflorescence - the tassel with staminateflowers
(ii) Female inflorescence - ear or cob with pistillate flowers. 
(iii) Unisexuality in maize is through selective abortion of stamens - in ear florets and Pistils - in tassel florets. 
(iv) A substitution of two single gene pairs 'ba' for barren plant and 'ts' - for tassel seed makes the differences between monoecious and dioecious (rare) maize plants. 
(v) The allele for barren plant (ba) when homozygous makes the stalk staminate by eliminating silk and ears. 
(vi) The allele for tassel seed (ts) transforms tassel into a pistillate flower. 
(vii) The development of both inflorescence is the result of selective abortion of pistils or stamens. 

(i) Multiple alleles are reported in Nicotiana which are responsible for self, incompatibility.
(ii) The gene for self-compatibility can be designated as S, which has allelic series S₁, S₂, S₃, S₄ and S₅.
(iii) The cross-fertilizing tobacco plants were not homozygous as S₁S₁ or S₂S₂.
(iv) But, were heterozygous as S₁S₂, S₃S₄, S₅S₆.
(v) When crosses were made between different S₁S₂ plants, pollen tube did not develop normally.
(vi) But effective pollen tube developed when crossing was made with other than S₁S₂.
(vii) When crosses were made between S₁S₂ (seed parents) S₂S₃ (pollen parents), two kinds of pollen tubes were distinguished.
(viii) Pollen grains carrying S₂ were not effective.
(ix) Pollen grains carrying S₃ were capable of fertilization.
(x) In a cross between S₁S₂ X S₃S₄, all the pollens were effective.
(xi) So four kinds of progeny resulted. S₁S₃/S₁S₄/S₂S₃/S₂S_4.

The widely accepted model of DNA recombination during crossing over is Holliday's hybrid DNA model.

Steps involved are:
1. Homologous DNA molecules are paired side by side with their duplicated copies of DNAs
2. One strand of both DNAs cut in one place by the enzyme endonuclease.
3. The cut strands cross and join the homologous strands forming the Holliday structure or Holliday junction.
4. The Holliday junction migrates away from the original site, a process called branch migration, as a result heteroduplex region is formed.
5. DNA strands may cut along through  the vertical (V) line or horizontal (H) line.
6. The vertical cut will result in heteroduplexes with recombinants.
7. The horizontal cut will result in heteroduplex with non recombinants.

Crossing over is a process that includes four stages.
1. Synapsis
(i) Pairing between two homologous chromosomes is initiated during sub stage zygotene stage of prophase I of meiosis I
(ii) Homologous chromosomes are aligned side by side resulting in the formation of bivalents. It also occurs in zygotene sub-stage of prophase I of meiosis I tetrads.This pairing phenomenon is called synapsis.
 2.Tetrad Formation
(i) Each homologous chromosome of a bivalent begin to form two identical sister chromatids.
(ii) They remain held together by a centromere at this stage each bivalent has four chromatids. This stage is called tetrad stage.
3. Cross Over
After tetrad formation, crossing over occurs in pachytene stage.
The non-sister chromatids of homologous pair make a contact at one or more points. These points of contact are called Chiasmata.
At chiasma, X-shaped structures are formed, where breaking and rejoining of two chromatids occur. This results in reciprocal exchange of equal and corresponding segments between them synapsis and chiasma formation are facilitated by protein filaments called synaptonemal complex.
4. Terminalisation
After crossing over, chiasma starts to move towards the terminal end of chromatids known as terminalisation. As a result, complete separation of homologous chromosomes occurs.

(i) Three point test cross.
(ii)

If we analyse the loci of two alleles at a time, starting with AB, since AB and ab are parental genotypes, the recombination will be Ab/aB.
The recombinant frequency for these two alleles can be calculated as follows.
\(RF = \frac{Total\ no.\ of \ recombination}{Total\ no.\ of\ progenies}\times100\)
\(\mathrm{RF}=\frac{114+5+4+116}{1200} \times 100 \)
\(\mathrm{RF}=\frac{239}{1200}=19.91 \)
For A and C loci, the recombinants are Ac / aC - RF will be as follows.
\(\mathrm{RF}=\frac{114+134+116+128}{1200} \times 100 \)
\(\mathrm{RF}=\frac{482}{1200}=40.1 \)
For B and C loci, the recombinants are Bc / bC- RF will be as follows.
\(R F=\frac{124+5+4+128}{1200} \times 100 \)
\(R F=\frac{261}{1200}=21.75 \)
All the loci are linked, because all the RF value is less than 50%. In this A - and C loci show highest RF value, they must be farthest apart. Therefore B locis must lie in between them.
The order of genes should be abc. A genetic map can be drawn as follows.
(iii)
 

(i) Complete Linkage
(ii)
 
(iii) Observed ratio 1:1 

Missense Mutation Non-sense Mutation The mutation where the codon for one amino acid is changed into a codon for another amino acid. The mutations where codon for one amino acid is changed into a termination or stop codon.

(i) Sutton and Boveri independently proposed the chromosome theory of inheritance.
(ii) Somatic cells of organisms are derived from the zygote by repeated mitotic division. These consist of two identical sets of chromosomes. One set is received from female parent and the other from male parent which constitute the homologous pair.
(iii) Chromosomes retain their structural uniqueness and individuality throughout the life cycle of an organism.
(iv) Each chromosome carries specific determiners or Mendelian factors which are now termed as genes.
(v) The behaviour of chromosomes during the gamete formation provides evidence to the fact that genes are located on chromosomes.