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Question

A tall violet flowered plant (TTVV) is crossed with a dwarf white flowered plant (ttvv). When the F1 plants are self pollinated, a total of 1600, F2 plants are produced. In the same experiment how many F2 plants show F1 phenotype?


A

900

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B

600

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C

300

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D

100

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Solution

The correct option is A

900


This is a typical dihybrid cross. Both parents are homozygous, so two different types of gametes will be formed- (TV) and (tv) respectively. The F1 will be heterozygous with the genotype TtVv. Four different gametes are formed by the F1 plants - (TV), (Tv), (tV), (tv). The phenotypes in the F2 progeny are observed in a 9:3:3:1 ratio.



As seen in the Punnett square above, 9/16 plants will have the dominant phenotype similar to F1 progeny. Therefore out of 1600 F2 plants, if 9/16 of them are of the F1 phenotype; 9/16 * 1600 = 900 plants would express this phenotype.


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