In a Mendelian dihybrid cross between a pea plant that has seeds with yellow color and round shape (RRYY) and another one that has seeds with green color and wrinkled shape (rryy), select the possible phenotypes obtained on self pollination after F1 generation.
(A) Round yellow
(B) Round green
(C) Wrinkled yellow
(D) Wrinkled brown
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
The cross RRYY × rryy gives an F1 that is RrYy, with round, yellow seeds. On self pollination, the F1 forms four kinds of gametes in equal numbers: RY, Ry, rY and ry. By independent assortment, these combine to give four phenotypes in the F2.
(A) Round yellow Correct. Any plant with at least one R and one Y has round, yellow seeds. This is the largest class, 9 of every 16.
(B) Round green Correct. A plant with at least one R but yy has round, green seeds. This is a new combination, 3 of every 16.
(C) Wrinkled yellow Correct. A plant that is rr with at least one Y has wrinkled, yellow seeds. This is the other new combination, 3 of every 16.
(D) Wrinkled brown Incorrect. The seed colour gene has only two alleles, Y for yellow and y for green. No brown seed can appear. The fourth class is wrinkled, green (rryy), 1 of every 16.
NCERT Extract:
Law of Independent Assortment –
Figure 4.7 Results of a dihybrid cross where the two parents differed in two pairs of contrasting traits: seed colour and seed shape