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3. In silkworms the hemolymph (functionally equivalent to blood) may be yellow or white. Fully developed larvae may be plain

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Answer #1

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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