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Authored by: Wol on Saturday, June 30 2012 @ 06:11 PM EDT |
Sorry, Celtic, but it would help if you actually READ what he wrote.
Field A contains pure GM. Field B contains pure non-GM.
ALL of field B is pollinated by field A.
Next year ALL of field B shows the GM trait despite the GM gene only being
present at the 50% level.
That's what the OP said, and it's true. They're all F1 GM. And that's why F1
doesn't breed true. You're correct for the NEXT generation where, if there is no
further contamination, only 75% will show the GM trait.
Cheers,
Wol[ Reply to This | Parent | # ]
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Authored by: Anonymous on Saturday, June 30 2012 @ 07:00 PM EDT |
Just do the analysis. Is the first generation is bb & aa,
then the children will be bb, ab, ba & aa. Dominant
means that 3/4 will express, but that doesn't change
the fact that half cary the recessive gene. So the second
generation needs paper to work out, but the numbers
are 16 bb, 16 ab, 16ba &16 aa. The proportion stays
the same.[ Reply to This | Parent | # ]
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