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Being born with six fingers is actually a dominant trait and the probability of the children would be 75% with six fingers and 25% with five fingers if both parents were heterozygous for that trait. If both parents were homozygous dominant for that trait then there is a 100% probability of the children being born with six fingers.
25%
both parents could be A heterozygous or one A heterozygous and the other O
3:4 or 75%
It is a 75% chance that the seeds will be round.
There is 50% that children will be of negative blood type if parents are heterozygous positive. 25% if one parent is homozygous and other heterozygous. 0% if both homozygous.
Yes, this can happen, if the alleles are Hetero for both parents (Rh+Rh-) heterozygous.
It can happen when both parents are a heterozygous
75% because the recessive and dominant genes are corresponding and in a Punnett square it takes over.
The probability is 3/4 or 75%. If both parents are heterozygous for the seed shape trait (e.g., Rr), there is a 50% chance that each parent will pass on the dominant allele (R) for round seeds to the offspring. The probability of inheriting the dominant allele from both parents and producing round seeds is therefore (1/2) x (1/2) = 1/4 or 25%. Since there are two possible ways to inherit the dominant allele (from either parent), the total probability is 2 x (1/4) = 1/2 or 50%.
Calculate the probability of a child having either sickle-cell anemia or cystic fibrosis if parents are each heterozygous for both.
Pp x Pp yields PP, Pp, Pp, pp. PP is the only genotype which will cause the phenotypic expression of the gene - symptoms of PKU. Therefore the probability is 1/4 or 25%