In pea plants, there is a dominant gene that causes some plants to be unnaturally tall. another dominant gene causes flowers to be purple instead of white. suppose a plant homozygous for the dominant form of both of these genes is mated to another plant that is homozygous for the recessive form of both genes. what fraction of the f2 generation will be tall and white? g

1/16

Explanation:

Let T represents the allele for unnatural tallness and P represents the allele for purpleness. The alternate form of both traits (tallness and whiteness) will be t and p respectively.

The genotype of a plant that is homozygous dominant for both traits will be TTPP while the one that is homozygous recessive for both trait will be ttpp.

If both are mated:                TTPP     x       ttpp

All the F1 offsprings will be unnaturally tall and purple with TtPp genotype

If two of the F1 offsprings are mated, the two traits will assort independently and produce the following phenotype at F2 according to Mendelian ratio:

Unnaturally tall, Purple flower colour     T_P_       9/16

Unnaturally tall, White flower colour       T_pp        3/16

Tall, Purple flower colour                          ttP_          3/16

Tall, White flower colour                            ttpp          1/16

The answer is 1/16 (see the attached image for Punnet's square analysis)

if you add everything up correctly you should get 69 i got this because 50 plus 19 is it