N a dihybrid cross for round and yellow seeds (RrYy x RrYy), what is the probability of having green and wrinkled seeds?
2 answers:
Answer:
1/16
Explanation:
See the Punnett square diagram to observe the result of the cross between the round and yellow plants (RrYy X RrYy).
Since round (R) is dominant over wrinkled (r) and yellow (Y) is dominant over y (green), the plant with genotype rryy would have wrinkled and green phenotype as these are recessive characters.
From the given cross in the Punnett square, we can observe that there is a total of 16 possible progenies and out of these only 1 has the wrinkled and green phenotype.
Therefore the probability is 1/16.
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Answer:
1/2 or 50%
Explanation:
This question involves a single gene coding for the number of digits per paw in cats or humans. The allele for polydactyly i.e six digits (P) is dominant over the allele for normal number of digits i.e 5 (p).
According to this question, If a man who is heterozygous (Pp) for this trait marries a woman with 5 fingers (pp), the following gametes will be produced by each parent:
Pp : P and p
pp : p and p
Using these gametes in a punnet square (see attached image), the following proportion of offsprings will be produced:
Pp, Pp, pp and pp
The offsprings with genotype Pp will have polydactyl, hence, the chances that they will have a child with polydactyly is 1/2 or 50%.
