Pea plant with true breeding nature for yellow and wrinkled seeds is crossed with another true breeding plant with green and round seeds. The $$F1individuals$$ are intercrossed and there is formation of $$240$$ pea plants in $$F2$$ generation. How many of $$theF2$$ plants have the genotype YY Rr

Question

Pea plant with true breeding nature for yellow and wrinkled seeds is crossed with another true breeding plant with green and round seeds. The $$F1individuals$$ are intercrossed and there is formation of $$240$$ pea plants in $$F2$$ generation. How many of $$theF2$$ plants have the genotype YY Rr

Infinity Learn · Accepted Answer

The probability of formation of $$F2$$ individuals with YY Rr genotype is product of probability of formation of YY genotype and Rr genotype i.$$e.14$$ × $$12$$ $$=$$ $$18$$. $$18$$ individuals of $$F2have$$ this genotype i.$$e.240$$ ×$$18$$ $$=$$ $$30$$

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Pea plant with true breeding nature for yellow and wrinkled seeds is crossed with another true breeding plant with green and round seeds. The $F_{1}$ individuals are intercrossed and there is formation of 240 pea plants in $F_{2}$ generation. How many of the $F_{2}$ plants have the genotype YY Rr

The probability of formation of $F_{2}$ individuals with YY Rr genotype is product of probability of formation of YY genotype and Rr genotype i.e. $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$ . $\frac{1}{8}$ individuals of $F_{2}$ have this genotype i.e. $240 \times \frac{1}{8} = 30$

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