Question

Dihybrid cross is made between RRYY (yellow round seed) and yyrr (wrinkled green seed) of pea. $F_2$ ratio amongst round yellow, round green, wrinkled yellow and wrinkled green would be
A. 9 : 3 : 3 : 1
B. 15 : 1 : 0 : 0
C. 9 : 3 : 4 : 0
D. 4 : 4 : 4 : 4

Answer 1

Hint: Mendel crossed pea plants differing in two characters (dihybrid cross) to verify the results of monohybrid crosses and it helped him to understand the inheritance of two pairs of genes at a time.

Complete answer:
A cross was made between a pure round yellow seeded pea plant (RRYY) with wrinkled green seeded pea plant (rryy). The yellow colour is dominant over green and round seed shape is dominant over wrinkled seed shape.

Gametes	RY	rY	Ry	ry
RY	RRYY	RRYY	RRYy	RrYy
rY	RrYY	rrYY	RrYy	rrYy
Ry	RRYy	RrYy	RRyy	Rryy
ry	RrYy	rrYy	Rryy	rryy

                                                         DIHYBRID CROSS
In $F_2$ generation mendel obtain 16 zygotes or 16 plants.

Phenotypic ratio :

Round yellow	Round green	Wrinkled yellow	Wrinkled green
9/16	3/16	3/16	1/16

1 RRYY (pure round, pure yellow)	2 RRYy (pure round, hybrid yellow)	1 RRyy (pure round, green)
2 RrYY (hybrid round, pure yellow)	4 RrYy (hybrid round, hybrid yellow)	2 Rryy (hybrid round, green)
1 rrYY (wrinkled, pure yellow)	2 rrYy (wrinkled, hybrid yellow)	1 rryy (wrinkled, green)

Genotypic ratio: 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 (9 types of genotypes)

So the correct answer is option A.

Note: Mendel found that in $F_1$ generation have all yellow and round seeds because yellow and round traits are dominant over green and wrinkled traits but when mendel self hybridized the $F_1$ generation he found that 3/4^th plants had yellow seeds and 1/4^th had green. This means yellow and green colour segregate in a 3 : 1 ratio.