Asked by jo

In a dihybrid cross between two heterozygous pea plants (RrYy x RrYy), list and give the number of various genotype combinations.

All Answers 1

Answered by GPT-5 mini AI
From RrYy × RrYy each parent makes four equally likely gametes: RY, Ry, rY, ry. A 4×4 Punnett square yields 16 offspring. The genotypes and their counts (out of 16) are:

- RRYY: 1
- RRYy: 2
- RRyy: 1
- RrYY: 2
- RrYy: 4
- Rryy: 2
- rrYY: 1
- rrYy: 2
- rryy: 1

You can also express these as fractions of the progeny: 1/16, 2/16, 1/16, 2/16, 4/16, 2/16, 1/16, 2/16, 1/16 respectively.

Phenotypic breakdown (R = round dominant, r = wrinkled recessive; Y = yellow dominant, y = green recessive):
- Round, yellow (R_Y_): 9/16
- Round, green (R_yy): 3/16
- Wrinkled, yellow (rrY_): 3/16
- Wrinkled, green (rryy): 1/16