The F1 progeny from crossing aa bb (1 g) with AA BB (3.4 g) plants will all have the genotype Aa Bb, resulting in a weight of 2.2 g due to additive effects from two dominant alleles at each locus.​

Problem Breakdown

Plant weight follows polygenic inheritance with two independently assorting loci (A/a and B/b), where recessive alleles (a, b) contribute 0 to weight beyond the base, and dominant alleles (A, B) add equal amounts. The base weight for aa bb is 1 g, while AA BB adds effects from four dominant alleles to reach 3.4 g. Each dominant allele contributes (3.4 – 1)/4 = 0.6 g.​

Step-by-Step Solution

• Identify genotypes: Parent 1 is aa bb (0 dominant alleles, weight = 1.0 g); Parent 2 is AA BB (4 dominant alleles, weight = 3.4 g).

• Calculate additive effect per dominant allele: Total increase = 2.4 g over 4 alleles, so 0.6 g each.

• F1 genotype: All progeny are Aa Bb (heterozygous at both loci, 2 dominant alleles total: one A and one B).

• F1 weight: Base 1.0 g + 2 × 0.6 g = 2.2 g (rounded to one decimal place).​

Why No Options?

This appears to be a fill-in-the-blank CSIR NET-style question without multiple choices, focusing on precise calculation rather than selecting from options. No “every option” evaluation applies, as confirmed by similar problems emphasizing direct computation.​

Introduction to Plant Weight Two Loci Additive Alleles

In plant genetics, traits like weight often show polygenic inheritance where multiple loci contribute additively. For CSIR NET aspirants, understanding how pairs of alleles at independently assorting loci (Aa and Bb) determine plant weight is crucial, especially when recessive alleles contribute nothing. This guide solves the cross between aa bb (1 g) and AA BB (3.4 g) plants, revealing the F1 progeny weight.​

Understanding Additive Effects in Polygenic Traits

Additive alleles mean each dominant (A or B) adds an equal increment to a base weight set by recessives. Here, aa bb provides the 1 g baseline with zero contribution from a or b. AA BB has four dominants, increasing weight by 2.4 g total (0.6 g per allele). F1 (Aa Bb) inherits one dominant per locus, adding exactly half the maximum increment.​

Detailed F1 Progeny Calculation

Cross aa bb × AA BB yields uniform Aa Bb F1.

• Dominant alleles in F1: 2 (A from one parent, B from the other).

• Effect: 1.0 + 2(0.6) = 2.2 g.
  This intermediate phenotype exemplifies no dominance or epistasis, pure additivity key for quantitative traits in plants.​

Relevance for CSIR NET Life Sciences

Such problems test polygenic concepts in genetics units, mirroring wheat kernel color or human height examples. Practice reinforces calculating genotypic values for F2 distributions too, vital for competitive exams.​