yellow, plain * white, moricaud
the ratio in the F2 is
yellow, Moricaud: yellow, plain: white, moricaud: white, plain=293:96:104:38=8.8:2.89:3.13:1.14 ( to get the ratio 1) find the total number of offsprings=531 2) divide the total nummber of offspring with 16, 16 is the total number of outcomes from the punnet square 531/16=33.2, 3) then divide number of each individuals with 33.2
293/33.2=8.8, 96/33.2=2.89,104/33.2=3.13, 38/33.2=1.14
this ratio is almost equal to 9:3:3:1 that is expected ratio of a dihybrid cross
large number of offsprings with dominant traits for characters are expeted in the F2, here large number of progenies are obtaned in the phenotypic class yello, moricaud means yellow and moricaud are dominant alleles so
there are 2 genes
gene 1 determines body colour pattern
moricaud- M is the dominant allele and m-plain is the recessive allele
gene 2 determines the body colour
Yellow -Y is the dominant allele and white-y is the recessive allele
so the first cross
yellow, plain * white, moricaud
YYww*yyMM
YyMm ( F1, yellow moricaud)
F1 cross
YyMm * YyMm
YM | Ym | yM | ym | |
YM | YYMM ( yellow, moricaud) | YYMm ( yellow, moricaud) | YyMM ( yellow, moricaud) | YyMm ( yellow, moricaud) |
Ym | YYMm ( yellow, moricaud) | YYmm ( yellow, plain) | YyMm ( yellow, moricaud) | Yymm yellow, plain) |
yM | YyMM ( yellow, moricaud) | YyMm ( yellow, moricaud) | yyMM ( white, moricaud) | yyMm( white, moricaud) |
ym | YyMm ( yellow, moricaud) | Yymm yellow, plain) | yyMm ( white, moricaud) | yymm (( white, plain) |
genotypes and phenotypes in the F2 is given in the punnet square.
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u tell me if these answers are correct please!??!!!
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