Search any question & find its solution
Question:
Answered & Verified by Expert
A pure homozygous pea plant with yellow round seeds is crossed with a pure homozygous pea plant with green wrinkled seeds. Their progeny produced is allowed to undergo self-pollination. It produces four different phenotypes. Each phenotype is an expression of different types of genotypes. Select the correct option from the table.
Options:
| Yellow round seeds | Green round seeds | Yellow wrinkled seeds | ||
| I. | Types of genotypes: | 9 | 3 | 3 |
| II. | Types of genotypes: | 4 | 2 | 2 |
| III. | Types of genotypes: | 3 | 3 | 3 |
| IV. | Types of genotypes: | 5 | 2 | 2 |
Solution:
2724 Upvotes
Verified Answer
The correct answer is:
II.
In the dihybrid cross, yellow round seeds are produced because of genotypes YYRR, YyRr, YyRR and YYRr (4 genotypes). Green Round seeds are produced by genotypes yyRR and yyRr (2 genotypes). Yellow wrinkled seeds are produced due to YYrr and Yyrr (2 genotypes).
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.