Question

A tobacco plant heterozygous for albinism is self-pollinated and 1200 seeds are subsequently germinated. How many seedlings would have the parental genotype?
a) 900
b) 600
c) 1200
d) 300

Answer 1

Hint: Mendel experimented with the inheritable traits of the pea plant and gave various theories related to the inheritance. He chooses the pea plant and its seven physically present characters to study. The control in the pollination is easy in pea plants making it suitable for the experiment.

Complete answer:
Albinism in the plant is the homozygous recessive character which is shown only in the recessive plant variety. The heterozygous for this character is a carrier and is unaffected. Let’s take the allele for the dominant albinism as ‘A’ and for the recessive let’s take it as ‘a’. So, for the heterozygous plant, it would be ‘Aa’ and for the dwarf plant, it would be ‘aa’.

A	a
AA	Aa	A
Aa	aa	a

When a cross of a single character is made a monohybrid, the cross is used. The heterozygous plant is crossed with itself (self-crossing). Their crossing yields progeny to be homozygous recessive, the heterozygous dominant, and the homozygous dominant.
Therefore, the genotypic and the phenotypic ratio obtained would be 1:2:1 and 3:1 respectively. The progeny obtained from this cross would yield 50% parental type and the rest would be recombinant type. Therefore, if 1200 plants were self-crossed then only 600 plants would develop the parental genotype.

So, the correct answer is option B.

Note: The punnett square is used in different types of crosses which are monohybrid cross where the only character is taken into the study, dihybrid cross where two characters are studied together, and the trihybrid cross in which three characters are studied together.