15. Body weight of rabbits is determined by pairs of alleles at two loci, ‘a’ and ‘b’ that are additive and equal in their effects. Rabbits with genotype a^– a^– b^– b^– have average 1 kg body weight, whereas individuals with genotype a⁺ a⁺ b⁺ b⁺ have animals that average 3.4 kg in weight. A male rabbit with a^– a^– b^– b^– crossed with a female of genotype a⁺ a⁺ b⁺ b⁺ What will be predicted average weight of F1 progeny of this cross?
(1) 2.2 kg         (2) 1.6 kg
(3) 1.2 kg         (4) 2.8 kg

The key to solving the problem lies in understanding additive gene action. The genotype a− a− b− b− corresponds to an average body weight of 1 kg and a+ a+ b+ b+ corresponds to 3.4 kg. Since the genes at two loci have equal and additive effects, the heterozygous genotype (F1 progeny from crossing a− a− b− b− with a+ a+ b+ b+) will have an average body weight that is the midpoint between these two extremes.

Calculation:

• Minimum weight (a− a− b− b−): 1 kg

• Maximum weight (a+ a+ b+ b+): 3.4 kg

• F1 genotype will be a+ a− b+ b− (heterozygous at both loci)

• Since effects are additive and equal, the F1 average weight is the average of these two weights: (1 + 3.4) / 2 = 2.2 kg

Thus, the predicted average weight of the F1 progeny is 2.2 kg.

Explanation of options:

1. 2.2 kg – Correct as it matches additive gene action midpoint.

2. 1.6 kg – Too low, does not fit midpoint under additive model.

3. 1.2 kg – Too low, very close to the lower extreme.

4. 2.8 kg – Too high, closer to the maximum but F1 heterozygous should be intermediate.

Introduction:
The body weight of rabbits can be genetically determined by additive effects of alleles at two loci. This article explains how to predict the average body weight of F1 progeny from a cross between rabbits with contrasting genotypes, using the genetic principle of additive gene action with equal effects at both loci.

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