Goal: to help you link cell division to Mendelian genetics; show with diagrams why/how Mendel’s dihybrid cross produces a ratio of 9:3:3:1 among F2 progeny. The genes for Mendel’s characters seed shape and seed color, are on different chromosomes. Each gene has two alleles. Starting, as Mendel did, with “true breeding” parents, show in writing and as part of a diagram, what their genotype is, the genotype of their gametes, the genotype of their progeny (the F1), the genotypes of their progeny’s gametes, and the genotypes of the F2. Include statements about the proportion of each kind of the genotypes you identify. To minimize the number of figures you should draw, I suggest that you illustrate the following stages: Parents genotypes (as I did in class, use one color for pollen donor (male function), another for plant fertilized by the pollen, different chromosomes different sizes/shapes) Anaphase I of meiosis in parents Anaphase II of meiosis in parents Gametes produced by meiosis in parents and their proportions. F1 genotypes Anaphase I of meiosis in F1 Anaphase II of meiosis in F2 Gametes produced by meiosis in F2 and their proportions. F2 Genotypes and their proportions; do this with letters only, not pictures. Solution To start with we have homozygotic parents : First parent: Round color seed (RR) and Yellow in color (YY) = Genotype : RRYY : Gametes RY and RY Second parent : wrinkled seed (rr) and green in colour (yy) = Genotype : rryy : Gametes ry and ry first offspring F1 = round and yellow RrYy Gamtes of F1 = RY, Ry, rY, ry Now crossing two individuals from F1 to obtain F2 generation now to see the proportion of all the phenotypes: Total number of F2 offspring Round and yellow = (RRYY) or (RrYy) or (RRYy) or (RrYY) = 9 Round and green = (RRyy) or (Rryy) = 3 Wrinkled and yellow = (rrYY) or (RRYy) = 3 Wrinkled and green = (rryy) = 1 Therefore the ratio is 9:3:3:1 in the F2 progeny.1st parent (RY)1st parent (RY)2nd parent (ry)RrYyRrYy2nd parent (ry)RrYyRrYy.

1. Goal: to help you link cell division to Mendelian genetics; show with diagrams why/how
Mendel’s dihybrid cross produces a ratio of 9:3:3:1 among F2 progeny. The genes for Mendel’s
characters seed shape and seed color, are on different chromosomes. Each gene has two alleles.
Starting, as Mendel did, with “true breeding” parents, show in writing and as part of a diagram,
what their genotype is, the genotype of their gametes, the genotype of their progeny (the F1), the
genotypes of their progeny’s gametes, and the genotypes of the F2. Include statements about the
proportion of each kind of the genotypes you identify. To minimize the number of figures you
should draw, I suggest that you illustrate the following stages: Parents genotypes (as I did in
class, use one color for pollen donor (male function), another for plant fertilized by the pollen,
different chromosomes different sizes/shapes) Anaphase I of meiosis in parents Anaphase II of
meiosis in parents Gametes produced by meiosis in parents and their proportions. F1 genotypes
Anaphase I of meiosis in F1 Anaphase II of meiosis in F2 Gametes produced by meiosis in F2
and their proportions. F2 Genotypes and their proportions; do this with letters only, not pictures.
Solution
To start with we have homozygotic parents :
First parent: Round color seed (RR) and Yellow in color (YY) = Genotype : RRYY : Gametes
RY and RY
Second parent : wrinkled seed (rr) and green in colour (yy) = Genotype : rryy : Gametes ry and
ry
first offspring F1 = round and yellow RrYy
Gamtes of F1 = RY, Ry, rY, ry
Now crossing two individuals from F1 to obtain F2 generation
now to see the proportion of all the phenotypes:
Total number of F2 offspring Round and yellow = (RRYY) or (RrYy) or (RRYy) or (RrYY) = 9
Round and green = (RRyy) or (Rryy) = 3
Wrinkled and yellow = (rrYY) or (RRYy) = 3
Wrinkled and green = (rryy) = 1
Therefore the ratio is 9:3:3:1 in the F2 progeny.1st parent (RY)1st parent (RY)2nd parent
(ry)RrYyRrYy2nd parent (ry)RrYyRrYy