A woman has a baby that is type B blood and she has type O blood. There are two men that could be the father, one has type A and
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Answer:
The probability of producing a round and green plant from this cross is 3/16
Explanation:
This question involves two distinct genes coding for seed shape and seed colour. In the seed shape gene, allele for round seeds (R) is dominant over allele for wrinked seeds (r), while in the seed colour gene, allele for yellow seed (Y) is dominant over allele for green seeds (y).
A truebreeding plant means that the plant possesses homozygous genotype. Hence, a truebreeding pea plant with round and green seeds will have genotype: RRyy while true-breeding pea plant with wrinkled and yellow seeds will have genotype: rrYY. These two parents will produce F1 offsprings with genotype: RrYy (heterozygous).
When the F1 offsprings are self-crossed (RrYy × RrYy), the following allelic combinations of gametes will be produced by each F1 parent: RY, Ry, rY, ry
Using these gametes in a punnet square (see attached image), 16 possible F2 offsprings with a phenotypic ratio 9:3:3:1 will be produced
9- Round yellow offsprings (R_Y_)
3- Round green offsprings (R_yy)
3- Wrinkled yellow offsprings (rrY_)
1- Wrinkled green offsprings (rryy)
