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.
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.
If both parents are heterozygous for brown eyes (Bb), the possible allele combinations for their children can be represented in a Punnett square: BB, Bb, Bb, and bb. This means there is a 1 in 4 chance, or a 25% probability, that their child will inherit the blue eye genotype (bb) and therefore have blue eyes.
In genetics, widow's peak is typically represented as a dominant trait (W), while a straight hairline is recessive (w). If both parents are heterozygous (Ww), their offspring can be represented by a Punnett square, which shows the possible genotypes: WW, Ww, Ww, and ww. Only the ww genotype results in a straight hairline, which occurs in 1 out of 4 possibilities. Therefore, the probability that two heterozygous parents will have a child with a straight hairline is 25%.
The probability that a given child will have PKU is 25%. This is because both parents are carriers of the recessive allele (heterozygous), so there is a 25% chance that they will each pass on the recessive allele, resulting in the child having PKU.
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.
To determine the probability of offspring being i barred females, we need information about the genetic makeup of the parents, specifically their genotypes related to the i barred trait. If both parents carry the i barred allele, we can use a Punnett square to analyze the potential combinations. The resulting probability will depend on the specific alleles present in the parents. Generally, if both parents are heterozygous for the i barred trait, the probability of producing i barred females could be calculated based on the ratios established in the Punnett square.