A variety of opium poppy (Papaver somniferum L.) having lacerated leaves was crossed with a variety that has normal leaves. All the F1 had lacerated leaves. Two F1 plants were interbred to produce the F2. Of the F2, 249 had lacerated leaves and 16 had normal leaves. Give genotypes for all the plants in the F1, and F2 generations. Explain how lacerate leaves are determined in the opium poppy. (Section 5.2)

The correct answer is-

F1 -  AaBb (lacerate)

F2 - A_B_; A_bb; and aaB_ (lacerate)

     - aabb (normal)

2. Two genes, with a dominant allele at either or both loci.

Explanation:

The given information gives THE following data:

Dominant: Lancerate leaves - AABB

Recessive: normal leaves - aabb

F1 has - all Lacerated leaves - AaBb

F2 by selfing F1:

         AB               Ab         aB         ab

AB    AABB        AABb      AaBB    AaBb

Ab     AABb       AAbb      AaBb     Aabb

aB     AaBB       AaBb       aaBB     aaBb

ab     AaBb       Aabb       aaBb      aabb

Here, 15/16 = lacerate which is 0.94 which is equal to the value of lacerate given in the question - 249/265 = 0.94

And normal 1/16 = 0.062 is almost same as 16/265 = 0.060

Thus, the genotypes of -

F1 -  AaBb (lacerate)

F2 - A_B_; A_bb; and aaB_ (lacerate)

     - aabb (normal)

dwadwadadaddadwadwdwdwad

complete the square on the right side of the equation;

-5(x + 0)^2 + 15

answer: d. y = -x^2 + 15