A heterozygous round and yellow seeded pea plantwere selfed and total 1600 seeds were obtained.Find out the number of plants develop from theseseeds, that are heterozygous for seed shape andhomozygous recessive for seed colour.
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150
Explanation:
The F2 ratio for the dihybrid cross between a heterozygous round and yellow pea plants is 9:3:3:1. The first dominant and second recessive traits will be round and green respectively. It is a recombinant trait. So, as per the ratio, the plants with such a trait will be:
The ratio of recombinant plants
Total Ratio X Total number of plants
3/16 ×800 = 150 Plants
Hence, the correct answer is '150'.
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