Explain why it's possible for two parents with a dominant trait like freckles to produce 10 children, all of whom do not have freckles. Use a punnett square to illustreat your answer.
2006-09-20 13:53:42 · 6 answers · asked by Chantal..carpe diem 3 in Education & Reference ➔ Homework Help
how do you make the square ?????????????
2006-09-20 14:02:54 · update #1
it is highly unlikely that any of the kids will not be freckless.
F =freckels f = no freckels
Parents – Ff x Ff
Genotypes of parents – Ff x Ff
Phenotypes of parents – (Ff) freckles and (Ff) freckels
Ff
FFFFf
fFfff
Genotypes of offspring – 25%FF : 50%Ff : 25%ff
Phenotypes of offspring – 75%freckless : 25%non freckless
there is a possibility of a mutation of mistakes while copying/replicating DNA
2006-09-22 03:25:54 · answer #1 · answered by ferrari_neville 1 · 0⤊ 0⤋
It is unlikely, although possible, that all children will have gotten 2 copies of the recessive allele (one from each parent), if each parent is heterozygous (has a freckle allele and a non-freckle allele).
Each parent is Ff, and all children are ff.
Parent's squares are both Ff, so the square is:
FF Ff
Ff ff
And the kids all got ff (circle ff on your sheet, I guess).
2006-09-20 13:56:46 · answer #2 · answered by sarahg 3 · 0⤊ 0⤋
the flying pig is correct.... what seems to trip people up is the probability issue... go back to beginning of statistics, the outcome of each event is independant of the other results. Each child had a 25% chance of the double recessive outcome...
2006-09-20 14:09:35 · answer #3 · answered by Karen 2 · 0⤊ 0⤋
One square will have both recessive genes, the bottom right-hand corner.
2006-09-20 13:56:07 · answer #4 · answered by gimme_danger 2 · 0⤊ 0⤋
its not possible for no children to have no freckles?
2006-09-20 13:56:01 · answer #5 · answered by lizzyb 4 · 0⤊ 1⤋
no
2006-09-20 14:00:26 · answer #6 · answered by Anonymous · 0⤊ 1⤋