In a dihybrid cross between individuals that are heterozygous for each trait, the four possible offspring phenotypes will be observed in the ratio of

***See attached pic***

Here's a made-up example to prove the point.

A = Tall plant

a = short plant

B = Big leaves

b = small leaves

(assuming 'A' is completely dominant to 'a', and 'B' is completely dominant to 'b')

If the parents are heterozygous for both traits then their genotypes will be "AaBb".

Make a 16-square Punnet Square and place all of the possible allele combinations for each parent gamete at the top (gametes are the sperm/egg and they have 1 copy of allele A/a and B/b --> if you've taken Algebra and you know how to FOIL something then do that to each genotype to find the possible allele combinations for each gamete).

Once you've created the Punnet Square you will see that:

9 plants = Tall and Big leaves (***see pic -blue squares***)

3 plants = Tall and small leaves (***see pic -purple squares***)

3 plants - short and Big leaves (***see pic -green squares***)

1 plant = short and small leaves (***see pic -orange square***)

RATION = 9:3:3:1

there are 35 white marbles

step-by-step explanation:

y- yellow marbles

4y + 5 = 45

subtract 5 from both sides

4y = 40

divide for from both sides

y = 10

plug in 10 to get 40 yellow marbles then subtract by 5 to get 35 marbles

thus there are only 10 yellow marbles

the answer would be 2, if crossing over doesn’t occur 2 out of 4 of the daughter cells will have that allele.

hope this : )