A cross between two inbred plants that had mean seed weights of 20 and 40 grams, respectively, produced an F1 generation with seeds weighing an average of 30 grams each. an F1 x F1 cross produced 1000 plants: 3 had seeds weighing 20 grams, 5 had seeds weighing 40 grams, and the others produced seeds with weights varying between these extremes. How many gene pairs are involved in determining seed weight differences in these plants?
This problem involves deciding carefully how the genotypes (gene arrangements) of different progeny can be produced.
It needs time and patience!!
Without actually giving you the solution think about a monohybrid cross.(Characteristics decided by a single pair of genes)
If the weight was decided by a single pair of genes WW or ww then
(Parent 1) P1 = WW (Parent 2) P2 = ww
and when P1 would be crossed with P2 then the F1 generation would segregate the genes in the ratio 3:1.
P1 = 40g P2 =20g F1 = 30g
The crossing of ...
This solution uses a step by step approach to apply logic in determining the number of gene pairs involved in a Mendelian problem. It relies on creating Punnet squares and looking at statistical ratios of the phenotypes produced.