Step 1: Identify the parental genotypes and F1 generation.
The parents are described as "pure round and yellow" and "wrinkled and green".
Let's assign alleles:
Round (R) is dominant over wrinkled (r)
Yellow (Y) is dominant over green (y)
So, the parental genotypes are:
Pure Round and Yellow: RRYY
Wrinkled and Green: rryy
The F1 generation (obtained by crossing RRYY x rryy) will have the genotype RrYy.
Step 2: Determine the F2 generation genotypes for the dihybrid cross.
The F2 generation is obtained by self-crossing the F1 generation (RrYy x RrYy). This is a standard Mendelian dihybrid cross. The genotypic ratio of the F2 generation is:
1 RRYY : 2 RRYy : 1 RRyy :
2 RrYY : 4 RrYy : 2 Rryy :
1 rrYY : 2 rrYy : 1 rryy
Step 3: Identify the genotypes that result in "wrinkled : yellow coloured seed" phenotype.
The question asks for the ratio of genotypes that correspond to the "wrinkled : yellow coloured seed" phenotype.
A wrinkled phenotype is represented by the homozygous recessive genotype 'rr'.
A yellow coloured seed phenotype is represented by 'YY' or 'Yy' (since yellow is dominant).
Combining these, the genotypes in the F2 generation that will exhibit the wrinkled yellow phenotype are:
rrYY (wrinkled and yellow)
rrYy (wrinkled and yellow)
Step 4: Determine the ratio of these genotypes.
From the F2 genotypic ratio listed in Step 2:
The proportion of rrYY is 1.
The proportion of rrYy is 2.
Therefore, the ratio of the genotypes (rrYY : rrYy) for wrinkled yellow coloured seeds is 1 : 2.
Step 5: Compare with the given options.
The calculated ratio of 1:2 matches option (2).
The final answer is $\boxed{\text{1 : 2}}$.