50%. Heterozygous means that there is two different traits inside of the gene. Therefore you have (aa) for the free earlobes and the other individual with attached (Aa). Drawing a Punnett square you get (aa) in two different spots, creating 50% probability.
Attached earlobes are recessive to free earlobes. What genotypic ratio is expected when an individual with attached earlobes mates with an individual heterozygous for free earlobes?
1:1
Mendel would call this a cross of Fa x aa which would yield Fa Fa AA AA or 2 attached and 2 heterozygous free. The ratio is then 1/1 for free to attached.
FF Ff or what ever letter symblizises the question
Blood typePossible genotypesAAA AOBBB BOBlood typePossible genotypesABABOOO
Yes it is your possibility if the parents were both heterozygous(having different alleles) or hybrid with Aa and Aa, the genotypic ratio would be 1:2:1 so if you put it in a punnet square there is a 25% chance of AA, 50% chance of Aa and 25% chance of aa.
punnet square
these nutz
The letters on the outside are the genotypes of two parents. Inside are the possible genotypes of possible offspring of the two
The possible genotypes HH and Hh. 50% homozygous for hanging earlobes (HH), and 50% heterozygous for hanging earlobes (Hh).
TT or Tt
I think you have the question backwards, "Why isn't it possible to have more phenotypes than genotypes?" There are always more or an equal number of genotypes relative to phenotypes. The phenotype for a simple dominant/recessive interaction (for example) T for tall and t for short where TT is tall, Tt is tall and tt is short has three genotypes and two phenotypes. If T and t are co-dominant then TT would be tall, Tt would be intermediate and tt would be short. (Three phenotypes and three genotypes.)
Either TT or Tt, where T stands for dominant gene for tallness and t for recessive gene.
Blood typePossible genotypesAAA AOBBB BOBlood typePossible genotypesABABOOO
Many possible genotypes, producing ,any possible phenotypes.
There are many types of dwarfism. Acondroplastic dwarfism is a dominant trait. Living individuals are heterozygous for the trait as a homozygous dominant individuals have substantial skeletal anomalies that result in death in infancy.
not possible. if the mother is normal she must have normal genes which are dominant. do you mean if athe mother is a carrier?
Yes it is your possibility if the parents were both heterozygous(having different alleles) or hybrid with Aa and Aa, the genotypic ratio would be 1:2:1 so if you put it in a punnet square there is a 25% chance of AA, 50% chance of Aa and 25% chance of aa.
A Punnett Square is a helpful device for predicting the proportions of possible genotypes
The only possible outcome is EeWw, which will express the dominant genes but carry the recessive ones. They get one chromosome from each parent, but since the parents all have matching chromsomes in this case then it doesn't matter which one they get. Since one parent has EE, E is the only one that can be passed on. Since the other has ee, they can only pass on e. Therefore, the child can only possible have Ee, as they get one from each parent.
The different forms of a gene are called alleles. In Mendelian genetics, a gene has a dominant allele and a recessive allele. The dominant allele masks the recessive allele if present. So there are two possible dominant genotypes: homozygous dominant, in which both dominant alleles are present; and heterozygous, in which one allele is dominant and the other allele is recessive. The only way to express a recessive trait is to have the homozygous recessive genotype.