Heterozygous round and yellow seeded pea plan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In a dihybrid cross involving heterozygous round and yellow seeded pea plants $(RrYy 	imes RrYy)$,the phenotypic ratio in the $F_2$ generation is

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(D) In a dihybrid cross involving heterozygous round and yellow seeded pea plants $(RrYy 	imes RrYy)$,the phenotypic ratio in the $F_2$ generation is $9:3:3:1$.Here,the traits are:Round $(R)$ is dominant,Wrinkled $(r)$ is recessive.Yellow $(Y)$ is dominant,Green $(y)$ is recessive.The question asks for seeds with the first dominant trait (Round) and the second recessive trait (Green),which corresponds to the genotype $R-yy$.The probability of obtaining Round and Green seeds is $\frac{3}{16}$.Given the total number of seeds is $800$,the number of Round and Green seeds is calculated as:$800 	imes \frac{3}{16} = 50 	imes 3 = 150$.

Heterozygous round and yellow seeded pea plants were selfed and a total of $800$ seeds were collected. What is the total number of seeds with the first dominant trait (round) and the second recessive trait (green)?

Similar Questions

When a cross is made between a tall plant with yellow seeds $(TtYy)$ and a tall plant with green seeds $(Ttyy)$,what proportions of phenotypes in the offspring could be expected to be:
$(a)$ Tall and green.
$(b)$ Dwarf and green.

Probability of genotype $TTrr$ in $F_{2}$-generation of a dihybrid cross is

In a dihybrid cross between individuals with genotypes $TtRr$,if the total number of offspring is $16$,identify the number of individuals with genotypes $TtRr$ and $TtRR$ among them.

Which genotype represents a true dihybrid condition?

In a plant with genotype $RrYy$,if self-pollination occurs,what will be the ratio of the following genotypes: $RRYY, RrYY, RRYy, RrYy$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module