Question

44. In Nicotiana tabacum, there are two independently-assorting gene loci that influence the appearance of the yellow burley character, i.e. A and B. When both loci are homozygous for their recessive alleles, i.e. aabb, the plant is yellow. When a dominant allele, i.e. A or B, is present at either locus, the plant is green. (a) If a cross were made between a homozygous green strain monosomic for one of these chromosomes (ABB) and a disomic yellow strain (aabb), what phenotypes would you expect among the monosomics and disomics of the F? (b) If the F, monosomics and disomics were inbred (mono x mono, di x di), what ratio of F2 phenotypes would you expect from each cross?

Answer 1

A -

AB B AB AaBb aBb ab Green Green

All the progenies will be green in color.

B- AaBb crossed with AaBb

	AB	aB	Ab	ab
AB	AABB	AaBB	AABb	AaBb
aB	AaBB	aaBB	AaBb	aaBb
Ab	AABb	AaBb	AAbb	Aabb
ab	AaBb	aaBb	Aabb	aabb

Out of 16 progenies, 15 have either A or B or both. So they will produce a green color.

Only one progeny has aabb genotype. So it will produce yellow color.

So ratio is 15:1

aBb crossed with aBb.

aBb crossed with aBb aB, ab, B and b aB, ab, B and b aBb Bb a BB aBb aaBb aaBB a BB aBb abb Bb aBb aaBb abb aabb

Out of 16 progenies, 12 have either B allele. So they will produce a green color.

Four progeny has aabb or bb or abb genotype. So it will produce yellow color.

So the ratio is 12:4 or 3:1