In peas, axial (A) flower position is dominant to terminal (a), and tall (L) is dominant to short (l). If a plant that is heterozygous for both traits is allowed to self-fertilize, how many of the offspring would also be heterozygous for both traits?

The correct answer is - 1/4 or 25%.

Explanation:

The parents have a heterozygous condition for both trait flower position and height of the pea plant. The representation of allele would be AaLl for both parents and each parent produce the same gametes that would be AL, Al, aL, and al.

Then the Punnett square will be :

         AL         Al          aL        al

AL     AALL     AALl    AaLL   AaLl

Al     AALl      AAll     AaLl    Aall

aL      AaLL     AaLl    aaLL    aaLl

al       AaLl     Aall       aaLl    aall

Out of 16 offspring 4 have heterozygous which means:

The heterozygous condition will be = 4/16 = 1/4

Thus, the correct answer is - 1/4

that scientist would be robert hooke