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# The pea plant with round- shaped seeds with the yellow color seed (RRYY) crossed with another pea plant showing wrinkled shape with green cotyledons (rryy). The results of progeny in the F2 generation are produced in the following number.(i) Heterozygous for both shape and color of seeds ___ number.(ii) Dominant for color, but recessive for shape ___ number.(iii) Homozygous for both shape and color of seeds ___ number.(iv) Heterozygous for seed shape only ___ number.(a) i - 2, ii - 4, iii - 3, iv - 8(b) i - 4, ii - 2, iii - 3, iv - 8(c) i - 3, ii - 4, iii - 8, iv - 2(d) i - 4, ii - 3, iii - 2, iv - 8

Last updated date: 13th Jun 2024
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Hint: When a cross is carried out between round- shaped yellow color seeds (RRYY) and wrinkled shape green color seeds (rryy) it is a dihybrid cross as two characters are taken at a single time in a single cross. According to Mendel's law of independent assortment, when two characters are observed in a single cross both the characters separate independently from one another.

When Mendel crossed yellow colored round shape seeds (RRYY) with green colored wrinkled shape seeds (rryy) he observed a phenotypic ratio of 9:3:3:1 (i.e., 9 round yellow seeds, 3 round green seeds, 3 yellow wrinkled seeds, and 1 green wrinkled seeds). In the F2 generation,
(i) heterozygous for both shape and color of seeds (RrYy) is 4 numbers.
(ii) dominant for color, but recessive for shape (rrYY/rrYy) is 3 numbers.
(iii) homozygous for both the shape and color of seeds (RRYY/rryy) is 2 numbers.
(iv) heterozygous for seed shape only (Rryy/RrYY) is 8 numbers

So, the correct answer is i - 4, ii - 2, iii - 3, and iv - 8.