The Mendelian Crosses summarized and illustrated with Punnett Squares.
Mendel performed a specific sequence of crosses. He started with pure breeding parents (ie. homozygous for a trait) and
crossed them. Each parent can produce only one type of gamete since it is homozygous for that trait. The outcome of
crossing the P generation is the F1 generation who will all be heterozygous for the trait. Since Mendel looked only at
characters that were dominant and recessive, the F1 always displays the dominant phenotype. Next, he crossed the F1’s
with each other (a self cross). Since all the F1’s are heterozygotes, they will all produce 2 types of gamete, one with the
dominant allele and one with the recessive allele. The F2 generation will have a mix of genotypes in specific ratio; 1
homozygous dominant, 2 heterozygotes, and 1 homozygous recessive. Since there is clear dominance, there will be 2
phenotypes in the ratio of 3 dominant: 1 recessive.
When writing out the genotypes, use the same letter for both the dominant and recessive alleles of the genotype. It makes
it easier to see what is going on.
Use the Punnett Square to lay out the crosses. The Punnett Square has two parts. First, it shows all of the gametes that
each parent can produce. These are written across the top of the columns or at the left end of each row. Second, the
Punnett Square shows all of the possible combinations of those gametes. These combinations ARE their offspring ie. the
next generation. I am going to show the pattern of the P, F1, and F2 generations in the following tables.
Assume that there is a character for bebopness. It takes 2 forms: big bebopper and little bebopper. We know that the allele
for big bebopper is dominant so we represent it with B. The allele for little bebopper is recessive, so we represent it with b.
We have two pure breeding strains. These will be the Parental generation. Their genotypes are BB and bb. When the BB
strain produces gametes (their gonads produce egg or sperm by meiosis), they can only make (B) gametes. When the bb
strain produces gametes, they can only make (b) gametes. Assuming that we cross a male BB with a female bb, the Punnett
Square would look like this. I will color the P generation’s gametes.
Parental Generation
Female
gametes
(b)
(b)
Male gametes
(B)
(B)
bB
bB
bB
bB
The F1 generation (the P generation’s offspring) is shown in the green part of the table. You will
see 4 green squares. Each is filled with the genotype of an F1 individual. Since each parent could
only make one type of gamete (B or b), all of the F1 have the same genotype. They are all
heterozygous, Bb. Since B is dominant over b, they all have the same phenotype, big bebopper.
Now let’s self cross the F1 generation. In order to do this, they must produce gametes (their
gonads produce egg or sperm by meiosis). All F1 are heterozygotes, Bb, so they will all produce
2 types of gamete: (B) and (b). The Punnett Square will look like this. The gametes produced by
the F1 generation are marked green.
F1 Generation
Female
gametes
(B)
(b)
Male gametes
(B)
(b)
BB
Bb
bB
bb
The F2 generation (the F1 generation’s offspring) is shown in the blue
part of the table. The blue squares show all the possible ways that the
gametes can combine to produce the F2. The F1 includes individuals with
3 genotypes: 1 homozygous dominant, BB; 2 heterozygotes, Bb; and 1
homozygous recessive, bb. Since B is dominant over b, there will be only
2 phenotypes. The individuals with the BB and Bb genotypes will have
the big bebopper phenotype. The individuals with the bb genotype will
have the little bebopper phenotype. Since 3 of the 4 offspring will be BB
or Bb and only 1 is bb, the phenotype ratio is 3 big bebopper: 1 little
bebopper. The F2 will always show this ratio of the different genotypes
and phenotypes.