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.
9:3:3:1 The probability of having both recessive traits is 1:16.
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.
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.
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.
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%.
1/8 or 12.5%
The probability is 50%. There are four probabilities: dominant homozygous, recessive homozygous, or heterozygous.
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.
Impossible. You can only be heterozygous or homozygous, not both.
25%
3:4 or 75%
9:3:3:1 The probability of having both recessive traits is 1:16.
It is a 75% chance that the seeds will be round.
both parents could be A heterozygous or one A heterozygous and the other O
false
The answer to this is 1 minus the probability that they will have 3 or fewer children. This would happen only if they had a boy as the first, second or third child. The probability they have a boy as first child is 0.5 The probability they have a boy as second is 0.25 The probability they have a boy as third is 0.125 Thus the total probability is 0.875 And so the probability they will have more than three children is 1-0.875 or 0.125
The individual probability that a child born will be female is 50% or 0.5.Using this we can calculate the probability that at least one of the children will be female by:calculating the probability that none of the children will be female and then subtracting this from 1.The probability that all the children are male is therefore 0.53 = 0.5 * 0.5 * 0.5 = 0.125.Thus the answer is 1 - 0.125 = 0.875 = 87.5%