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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.
Wiki User
∙ 12y ago
The probability that a child will be heterozygous depends on the parent's genotypes. If one parent is heterozygous (Aa) and the other is also heterozygous (Aa), then there is a 50% chance (1 out of 2 possible genotypes) that their child will be heterozygous. If both parents are homozygous (AA or aa), then all their children will be heterozygous.
Wiki User
∙ 12y agoIt will depend on the genotype of the parents.
If both parents have the same genotype, say BB - then all of their children should also be BB.
If both parents are heterozygous, Bb - then 50% of their children should also be heterozygous (Bb), 25% BB and 25% bb.
If one parent is BB and the other bb - then all of their children will be heterozygous (Bb).
Wiki User
∙ 11y agoIf the trait for six fingers is dominant then the odds are 3/4, if the trait is recessive then the odds are 1/4
Wiki User
∙ 13y ago50%
Wiki User
∙ 12y ago1/2
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What are the ratios of a heterozygous dihybrid cross and what is the probability of having the recessive trait show?
9:3:3:1 The probability of having both recessive traits is 1:16.
What is the probability a woman heterozygous for an x-linked trait will have a son with a genetic disorder if the genetic disorder is recessive?
The probability would be 0.5 or 50%. A heterozygous woman will pass on the X chromosome with the recessive allele to 50% of her sons, and since the disorder is recessive, the son would only have the disorder if the X chromosome with the recessive allele is inherited from the mother.
What is the probability that the next two children will both be affected with galactosemia?
The probability that a child is affected with galactosemia is 1/40,000. The probability that both children are affected would be (1/40,000) * (1/40,000) = 1/1,600,000,000.
If both heterozygous parents are carriers of a recessive allele for phenylketonuria the probability that a given child of these parents will have PKU is?
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.
Galactosemia is a recessive human disease A woman and her husband are both heterozygous for the glactosemia gene She has identical twins What is the probability that they are both girls and have ga?
The probability both twins are girls is 0.25 (0.5 * 0.5). Each girl will inherit half of the mother's genes, so there is a 0.25 chance they both inherit the galactosemia gene from their heterozygous mother. Therefore, the probability that they are both girls and have galactosemia is 0.25 * 0.25 = 0.0625 or 6.25%.
What is a probability of 2 Rh plus heterozygous have 4 Rh- children?
1/8 or 12.5%
What is the probability of 2 heterozygous individuals producing a heterozygous recessive offspring?
The probability is 50%. There are four probabilities: dominant homozygous, recessive homozygous, or heterozygous.
What is the probability of offspring for 2 heterozygous dogs?
The probability of offspring for two heterozygous dogs (Aa x Aa) is 25% homozygous dominant (AA), 50% heterozygous (Aa), and 25% homozygous recessive (aa) based on Mendelian genetics principles.
What is the probability of a heterhomozygous offspring?
Impossible. You can only be heterozygous or homozygous, not both.
What is the probability of 2 heterozygous individuals producing a homozygous recessive offspring?
25%
What is the probability of obtaining a dominant phenotype from self-fertilization of a heterozygous individual is?
The probability of obtaining a dominant phenotype from self-fertilization of a heterozygous individual is 75%. This is because in a heterozygous individual, there is a 50% chance of passing on the dominant allele and a 50% chance of passing on the recessive allele. With self-fertilization, the possible combinations are: 1 dominant allele (25%), 2 dominant alleles (50%), and 1 recessive allele (25%). Dominant phenotype will be expressed if there are one or more dominant alleles present.
What was the probability that two heterozygous parents would have an offspring that produced round seeds?
If both parents are heterozygous for seed shape (Rr), their offspring would have a 75% chance of producing round seeds (3 out of 4 possible combinations), assuming round seeds (R) are dominant over wrinkled seeds (r).
What was the probability that two heterozygous parents would have an offspring that produced rounds seeds?
3:4 or 75%
What are the ratios of a heterozygous dihybrid cross and what is the probability of having the recessive trait show?
9:3:3:1 The probability of having both recessive traits is 1:16.
What is probability that two heterozygous parents would have an offspring that produced round seeds?
It is a 75% chance that the seeds will be round.
If children with blood group O and a what could be the blood group of parents?
both parents could be A heterozygous or one A heterozygous and the other O
Is The probability that a gamete produced by a pea plant heterozygous for stem height Tt will contain the recessive allele 100 percent?
false
