Question

In Garden Pea, yellow cotyledons are dominant over green, and a round shape of the seed is dominant over the wrinkled. When a plant with yellow and round seeds is crossed with a plant having yellow and wrinkled seeds, the progeny showed segregation for all four characters. The probability of having green round seeds in the progeny of this cross is:
A) 1/8
B) 1/16
C) 1/4
D) 3/16

Answer 1

In Mendelian genetics for unlinked genes the dominant gene will express the phenotype even in the heterozygous condition, however, the recessive gene will express only in the homozygous condition. Since both parents were phenotypically yellow, but produced green seeds, they would both be heterozygous.

Complete answer:
The parent plants would be genotypically YySs and Yyss as to produce wrinkled seeds in the progeny, the round plant would be heterozygous. The number of offspring having green round seeds would be 2 out 16 or 1/8. The correct answer is option A.

Option B shows the typical ration for a dihybrid cross, where the homozygous recessive will show both the recessive phenotypes, in this case green wrinkled. Option B is incorrect.

Option C and D also apply to a typical dihybrid cross. Giving the typical ratios for the dominant phenotypes and for progeny expressing one dominant and one recessive phenotype. Option C and D are incorrect. This cross would be found if the parents were homozygous for both traits: yellow round and green wrinkled. In this cross, the likely parental genotypes would be YySs and YYss, given that combinations were produced despite both being genotypically yellow. The ratio of offspring combinations will also differ from a typical dihybrid cross between two parents heterozygous for both characters, or homozygous for both.

Hence the correct answer is option ‘A’.

Note: Such crosses are calculated using Punnett squares, where the possible gamete combinations for each parent are arranged on the top and one side of the square, and then the various combinations are produced.